Calculating Motion in One Dimension - AP Physics C: Electricity and Magnetism

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Question

A man runs with a velocity described by the function below.

What is the function for his acceleration?

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Answer

The function for acceleration is the derivative of the function for velocity.

$a(t)= $\frac{dv(t)}{dt}$$

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Question

A man runs with a velocity as described by the function below.

How far does he travel in 1 minute?

Tap to reveal answer

Answer

Distance is given by the integral of a velocity function. For this question, we will need to integrate over the interval of 0s to 60s.

$x(t) = \int_${0}^{60}$v(t),dt$

$x(60)=\int_${0}^{60}$(8t+4),dt = $(4(60)^2$$+4(60))-(4(0)^2$+4(0))$

$x(60)=4(3600)+240 = 14640\ m$

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Question

A particle traveling in a straight line accelerates uniformly from rest to $50:\up$\text{cm/s}$$ in $5:\up$\text{s}$$ and then continues at constant speed for an additional for an additional $3:\up$\text{s}$$ . What is the total distance traveled by the particle during the $8:\up$\text{s}$$ ?

Tap to reveal answer

Answer

First off we have to convert $50:\up$\text{cm/s}$$ to meters per second.

$\frac{50:\uptext{cm}$}{s} \cdot $\frac{1:\uptext{m}$}{100\up$\text{cm}$} = 0.5:\up$\text{m/s}$

Next we have to calculate the distance the object traveled the first 5 seconds, when it was starting from rest. We are given time, initial speed to be . The acceleration of the object at this time can be calculated using:

, substituting the values, we get:

$0.5:\up$\text{m/s}$ = a(5s), a = $0.1:m/s^2$$

Next, we use the distance equation to find the distance in the first 5 seconds:

$x_1 = v_0t + $$\frac{1}{2}$at^2$$

because the initial speed is .

If we plug in , , we get: $x_1 = 1.25:\up$\text{m}$$

Next we have to find the distance the the object travels at constant speed for 3 seconds. We can use the equation:

in this case is not equal to , it is equal to $0.5:\up$\text{m}$$ and $t=3:\up$\text{s}$$

Plugging in the equation, we get

Adding and we get the total distance to be $2.75:\up$\text{m}$$

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Question

A man runs with a velocity described by the function below.

What is the function for his acceleration?

Tap to reveal answer

Answer

The function for acceleration is the derivative of the function for velocity.

$a(t)= $\frac{dv(t)}{dt}$$

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Question

A man runs with a velocity as described by the function below.

How far does he travel in 1 minute?

Tap to reveal answer

Answer

Distance is given by the integral of a velocity function. For this question, we will need to integrate over the interval of 0s to 60s.

$x(t) = \int_${0}^{60}$v(t),dt$

$x(60)=\int_${0}^{60}$(8t+4),dt = $(4(60)^2$$+4(60))-(4(0)^2$+4(0))$

$x(60)=4(3600)+240 = 14640\ m$

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Question

A particle traveling in a straight line accelerates uniformly from rest to $50:\up$\text{cm/s}$$ in $5:\up$\text{s}$$ and then continues at constant speed for an additional for an additional $3:\up$\text{s}$$ . What is the total distance traveled by the particle during the $8:\up$\text{s}$$ ?

Tap to reveal answer

Answer

First off we have to convert $50:\up$\text{cm/s}$$ to meters per second.

$\frac{50:\uptext{cm}$}{s} \cdot $\frac{1:\uptext{m}$}{100\up$\text{cm}$} = 0.5:\up$\text{m/s}$

Next we have to calculate the distance the object traveled the first 5 seconds, when it was starting from rest. We are given time, initial speed to be . The acceleration of the object at this time can be calculated using:

, substituting the values, we get:

$0.5:\up$\text{m/s}$ = a(5s), a = $0.1:m/s^2$$

Next, we use the distance equation to find the distance in the first 5 seconds:

$x_1 = v_0t + $$\frac{1}{2}$at^2$$

because the initial speed is .

If we plug in , , we get: $x_1 = 1.25:\up$\text{m}$$

Next we have to find the distance the the object travels at constant speed for 3 seconds. We can use the equation:

in this case is not equal to , it is equal to $0.5:\up$\text{m}$$ and $t=3:\up$\text{s}$$

Plugging in the equation, we get

Adding and we get the total distance to be $2.75:\up$\text{m}$$

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Question

A man runs with a velocity described by the function below.

What is the function for his acceleration?

Tap to reveal answer

Answer

The function for acceleration is the derivative of the function for velocity.

$a(t)= $\frac{dv(t)}{dt}$$

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Question

A man runs with a velocity as described by the function below.

How far does he travel in 1 minute?

Tap to reveal answer

Answer

Distance is given by the integral of a velocity function. For this question, we will need to integrate over the interval of 0s to 60s.

$x(t) = \int_${0}^{60}$v(t),dt$

$x(60)=\int_${0}^{60}$(8t+4),dt = $(4(60)^2$$+4(60))-(4(0)^2$+4(0))$

$x(60)=4(3600)+240 = 14640\ m$

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Question

A particle traveling in a straight line accelerates uniformly from rest to $50:\up$\text{cm/s}$$ in $5:\up$\text{s}$$ and then continues at constant speed for an additional for an additional $3:\up$\text{s}$$ . What is the total distance traveled by the particle during the $8:\up$\text{s}$$ ?

Tap to reveal answer

Answer

First off we have to convert $50:\up$\text{cm/s}$$ to meters per second.

$\frac{50:\uptext{cm}$}{s} \cdot $\frac{1:\uptext{m}$}{100\up$\text{cm}$} = 0.5:\up$\text{m/s}$

Next we have to calculate the distance the object traveled the first 5 seconds, when it was starting from rest. We are given time, initial speed to be . The acceleration of the object at this time can be calculated using:

, substituting the values, we get:

$0.5:\up$\text{m/s}$ = a(5s), a = $0.1:m/s^2$$

Next, we use the distance equation to find the distance in the first 5 seconds:

$x_1 = v_0t + $$\frac{1}{2}$at^2$$

because the initial speed is .

If we plug in , , we get: $x_1 = 1.25:\up$\text{m}$$

Next we have to find the distance the the object travels at constant speed for 3 seconds. We can use the equation:

in this case is not equal to , it is equal to $0.5:\up$\text{m}$$ and $t=3:\up$\text{s}$$

Plugging in the equation, we get

Adding and we get the total distance to be $2.75:\up$\text{m}$$

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Question

A man runs with a velocity described by the function below.

What is the function for his acceleration?

Tap to reveal answer

Answer

The function for acceleration is the derivative of the function for velocity.

$a(t)= $\frac{dv(t)}{dt}$$

← Didn't Know|Knew It →

Question

A man runs with a velocity as described by the function below.

How far does he travel in 1 minute?

Tap to reveal answer

Answer

Distance is given by the integral of a velocity function. For this question, we will need to integrate over the interval of 0s to 60s.

$x(t) = \int_${0}^{60}$v(t),dt$

$x(60)=\int_${0}^{60}$(8t+4),dt = $(4(60)^2$$+4(60))-(4(0)^2$+4(0))$

$x(60)=4(3600)+240 = 14640\ m$

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Question

A particle traveling in a straight line accelerates uniformly from rest to $50:\up$\text{cm/s}$$ in $5:\up$\text{s}$$ and then continues at constant speed for an additional for an additional $3:\up$\text{s}$$ . What is the total distance traveled by the particle during the $8:\up$\text{s}$$ ?

Tap to reveal answer

Answer

First off we have to convert $50:\up$\text{cm/s}$$ to meters per second.

$\frac{50:\uptext{cm}$}{s} \cdot $\frac{1:\uptext{m}$}{100\up$\text{cm}$} = 0.5:\up$\text{m/s}$

Next we have to calculate the distance the object traveled the first 5 seconds, when it was starting from rest. We are given time, initial speed to be . The acceleration of the object at this time can be calculated using:

, substituting the values, we get:

$0.5:\up$\text{m/s}$ = a(5s), a = $0.1:m/s^2$$

Next, we use the distance equation to find the distance in the first 5 seconds:

$x_1 = v_0t + $$\frac{1}{2}$at^2$$

because the initial speed is .

If we plug in , , we get: $x_1 = 1.25:\up$\text{m}$$

Next we have to find the distance the the object travels at constant speed for 3 seconds. We can use the equation:

in this case is not equal to , it is equal to $0.5:\up$\text{m}$$ and $t=3:\up$\text{s}$$

Plugging in the equation, we get

Adding and we get the total distance to be $2.75:\up$\text{m}$$

← Didn't Know|Knew It →

Question

A man runs with a velocity described by the function below.

What is the function for his acceleration?

Tap to reveal answer

Answer

The function for acceleration is the derivative of the function for velocity.

$a(t)= $\frac{dv(t)}{dt}$$

← Didn't Know|Knew It →

Question

A man runs with a velocity as described by the function below.

How far does he travel in 1 minute?

Tap to reveal answer

Answer

Distance is given by the integral of a velocity function. For this question, we will need to integrate over the interval of 0s to 60s.

$x(t) = \int_${0}^{60}$v(t),dt$

$x(60)=\int_${0}^{60}$(8t+4),dt = $(4(60)^2$$+4(60))-(4(0)^2$+4(0))$

$x(60)=4(3600)+240 = 14640\ m$

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Question

A particle traveling in a straight line accelerates uniformly from rest to $50:\up$\text{cm/s}$$ in $5:\up$\text{s}$$ and then continues at constant speed for an additional for an additional $3:\up$\text{s}$$ . What is the total distance traveled by the particle during the $8:\up$\text{s}$$ ?

Tap to reveal answer

Answer

First off we have to convert $50:\up$\text{cm/s}$$ to meters per second.

$\frac{50:\uptext{cm}$}{s} \cdot $\frac{1:\uptext{m}$}{100\up$\text{cm}$} = 0.5:\up$\text{m/s}$

Next we have to calculate the distance the object traveled the first 5 seconds, when it was starting from rest. We are given time, initial speed to be . The acceleration of the object at this time can be calculated using:

, substituting the values, we get:

$0.5:\up$\text{m/s}$ = a(5s), a = $0.1:m/s^2$$

Next, we use the distance equation to find the distance in the first 5 seconds:

$x_1 = v_0t + $$\frac{1}{2}$at^2$$

because the initial speed is .

If we plug in , , we get: $x_1 = 1.25:\up$\text{m}$$

Next we have to find the distance the the object travels at constant speed for 3 seconds. We can use the equation:

in this case is not equal to , it is equal to $0.5:\up$\text{m}$$ and $t=3:\up$\text{s}$$

Plugging in the equation, we get

Adding and we get the total distance to be $2.75:\up$\text{m}$$

← Didn't Know|Knew It →

Question

A man runs with a velocity described by the function below.

What is the function for his acceleration?

Tap to reveal answer

Answer

The function for acceleration is the derivative of the function for velocity.

$a(t)= $\frac{dv(t)}{dt}$$

← Didn't Know|Knew It →

Question

A man runs with a velocity as described by the function below.

How far does he travel in 1 minute?

Tap to reveal answer

Answer

Distance is given by the integral of a velocity function. For this question, we will need to integrate over the interval of 0s to 60s.

$x(t) = \int_${0}^{60}$v(t),dt$

$x(60)=\int_${0}^{60}$(8t+4),dt = $(4(60)^2$$+4(60))-(4(0)^2$+4(0))$

$x(60)=4(3600)+240 = 14640\ m$

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Question

A particle traveling in a straight line accelerates uniformly from rest to $50:\up$\text{cm/s}$$ in $5:\up$\text{s}$$ and then continues at constant speed for an additional for an additional $3:\up$\text{s}$$ . What is the total distance traveled by the particle during the $8:\up$\text{s}$$ ?

Tap to reveal answer

Answer

First off we have to convert $50:\up$\text{cm/s}$$ to meters per second.

$\frac{50:\uptext{cm}$}{s} \cdot $\frac{1:\uptext{m}$}{100\up$\text{cm}$} = 0.5:\up$\text{m/s}$

Next we have to calculate the distance the object traveled the first 5 seconds, when it was starting from rest. We are given time, initial speed to be . The acceleration of the object at this time can be calculated using:

, substituting the values, we get:

$0.5:\up$\text{m/s}$ = a(5s), a = $0.1:m/s^2$$

Next, we use the distance equation to find the distance in the first 5 seconds:

$x_1 = v_0t + $$\frac{1}{2}$at^2$$

because the initial speed is .

If we plug in , , we get: $x_1 = 1.25:\up$\text{m}$$

Next we have to find the distance the the object travels at constant speed for 3 seconds. We can use the equation:

in this case is not equal to , it is equal to $0.5:\up$\text{m}$$ and $t=3:\up$\text{s}$$

Plugging in the equation, we get

Adding and we get the total distance to be $2.75:\up$\text{m}$$

← Didn't Know|Knew It →

Question

A man runs with a velocity described by the function below.

What is the function for his acceleration?

Tap to reveal answer

Answer

The function for acceleration is the derivative of the function for velocity.

$a(t)= $\frac{dv(t)}{dt}$$

← Didn't Know|Knew It →

Question

A man runs with a velocity as described by the function below.

How far does he travel in 1 minute?

Tap to reveal answer

Answer

Distance is given by the integral of a velocity function. For this question, we will need to integrate over the interval of 0s to 60s.

$x(t) = \int_${0}^{60}$v(t),dt$

$x(60)=\int_${0}^{60}$(8t+4),dt = $(4(60)^2$$+4(60))-(4(0)^2$+4(0))$

$x(60)=4(3600)+240 = 14640\ m$

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Knew It

Calculating Motion in One Dimension - AP Physics C: Electricity and Magnetism

A man runs with a velocity described by the function below. What is the function for his acceleration?

The function for acceleration is the derivative of the function for velocity.

A man runs with a velocity as described by the function below. How far does he travel in 1 minute?

Distance is given by the integral of a velocity function. For this question, we will need to integrate over the interval of 0s to 60s.

A particle traveling in a straight line accelerates uniformly from rest to in and then continues at constant speed for an additional for an additional . What is the total distance traveled by the particle during the ?

A man runs with a velocity described by the function below. What is the function for his acceleration?

The function for acceleration is the derivative of the function for velocity.

A man runs with a velocity as described by the function below. How far does he travel in 1 minute?

Distance is given by the integral of a velocity function. For this question, we will need to integrate over the interval of 0s to 60s.

A particle traveling in a straight line accelerates uniformly from rest to in and then continues at constant speed for an additional for an additional . What is the total distance traveled by the particle during the ?

A man runs with a velocity described by the function below. What is the function for his acceleration?

The function for acceleration is the derivative of the function for velocity.

A man runs with a velocity as described by the function below. How far does he travel in 1 minute?

Distance is given by the integral of a velocity function. For this question, we will need to integrate over the interval of 0s to 60s.

A particle traveling in a straight line accelerates uniformly from rest to in and then continues at constant speed for an additional for an additional . What is the total distance traveled by the particle during the ?

A man runs with a velocity described by the function below. What is the function for his acceleration?

The function for acceleration is the derivative of the function for velocity.

A man runs with a velocity as described by the function below. How far does he travel in 1 minute?

Distance is given by the integral of a velocity function. For this question, we will need to integrate over the interval of 0s to 60s.

A particle traveling in a straight line accelerates uniformly from rest to in and then continues at constant speed for an additional for an additional . What is the total distance traveled by the particle during the ?

A man runs with a velocity described by the function below. What is the function for his acceleration?

The function for acceleration is the derivative of the function for velocity.

A man runs with a velocity as described by the function below. How far does he travel in 1 minute?

Distance is given by the integral of a velocity function. For this question, we will need to integrate over the interval of 0s to 60s.

A particle traveling in a straight line accelerates uniformly from rest to in and then continues at constant speed for an additional for an additional . What is the total distance traveled by the particle during the ?

A man runs with a velocity described by the function below. What is the function for his acceleration?

The function for acceleration is the derivative of the function for velocity.

A man runs with a velocity as described by the function below. How far does he travel in 1 minute?

Distance is given by the integral of a velocity function. For this question, we will need to integrate over the interval of 0s to 60s.

A particle traveling in a straight line accelerates uniformly from rest to in and then continues at constant speed for an additional for an additional . What is the total distance traveled by the particle during the ?

A man runs with a velocity described by the function below. What is the function for his acceleration?

The function for acceleration is the derivative of the function for velocity.

A man runs with a velocity as described by the function below. How far does he travel in 1 minute?

Distance is given by the integral of a velocity function. For this question, we will need to integrate over the interval of 0s to 60s.

A man runs with a velocity described by the function below.

What is the function for his acceleration?

The function for acceleration is the derivative of the function for velocity.

$a(t)= $\frac{dv(t)}{dt}$$

A man runs with a velocity as described by the function below.

How far does he travel in 1 minute?

Distance is given by the integral of a velocity function. For this question, we will need to integrate over the interval of 0s to 60s.

$x(t) = \int_${0}^{60}$v(t),dt$

$x(60)=\int_${0}^{60}$(8t+4),dt = $(4(60)^2$$+4(60))-(4(0)^2$+4(0))$

$x(60)=4(3600)+240 = 14640\ m$

A particle traveling in a straight line accelerates uniformly from rest to $50:\up$\text{cm/s}$$ in $5:\up$\text{s}$$ and then continues at constant speed for an additional for an additional $3:\up$\text{s}$$ . What is the total distance traveled by the particle during the $8:\up$\text{s}$$ ?

A man runs with a velocity described by the function below.

What is the function for his acceleration?

The function for acceleration is the derivative of the function for velocity.

$a(t)= $\frac{dv(t)}{dt}$$

A man runs with a velocity as described by the function below.

How far does he travel in 1 minute?

Distance is given by the integral of a velocity function. For this question, we will need to integrate over the interval of 0s to 60s.

$x(t) = \int_${0}^{60}$v(t),dt$

$x(60)=\int_${0}^{60}$(8t+4),dt = $(4(60)^2$$+4(60))-(4(0)^2$+4(0))$

$x(60)=4(3600)+240 = 14640\ m$

A particle traveling in a straight line accelerates uniformly from rest to $50:\up$\text{cm/s}$$ in $5:\up$\text{s}$$ and then continues at constant speed for an additional for an additional $3:\up$\text{s}$$ . What is the total distance traveled by the particle during the $8:\up$\text{s}$$ ?

A man runs with a velocity described by the function below.

What is the function for his acceleration?

The function for acceleration is the derivative of the function for velocity.

$a(t)= $\frac{dv(t)}{dt}$$

A man runs with a velocity as described by the function below.

How far does he travel in 1 minute?

Distance is given by the integral of a velocity function. For this question, we will need to integrate over the interval of 0s to 60s.

$x(t) = \int_${0}^{60}$v(t),dt$

$x(60)=\int_${0}^{60}$(8t+4),dt = $(4(60)^2$$+4(60))-(4(0)^2$+4(0))$

$x(60)=4(3600)+240 = 14640\ m$

A particle traveling in a straight line accelerates uniformly from rest to $50:\up$\text{cm/s}$$ in $5:\up$\text{s}$$ and then continues at constant speed for an additional for an additional $3:\up$\text{s}$$ . What is the total distance traveled by the particle during the $8:\up$\text{s}$$ ?

A man runs with a velocity described by the function below.

What is the function for his acceleration?

The function for acceleration is the derivative of the function for velocity.

$a(t)= $\frac{dv(t)}{dt}$$

A man runs with a velocity as described by the function below.

How far does he travel in 1 minute?

Distance is given by the integral of a velocity function. For this question, we will need to integrate over the interval of 0s to 60s.

$x(t) = \int_${0}^{60}$v(t),dt$

$x(60)=\int_${0}^{60}$(8t+4),dt = $(4(60)^2$$+4(60))-(4(0)^2$+4(0))$

$x(60)=4(3600)+240 = 14640\ m$

A particle traveling in a straight line accelerates uniformly from rest to $50:\up$\text{cm/s}$$ in $5:\up$\text{s}$$ and then continues at constant speed for an additional for an additional $3:\up$\text{s}$$ . What is the total distance traveled by the particle during the $8:\up$\text{s}$$ ?

A man runs with a velocity described by the function below.

What is the function for his acceleration?

The function for acceleration is the derivative of the function for velocity.

$a(t)= $\frac{dv(t)}{dt}$$

A man runs with a velocity as described by the function below.

How far does he travel in 1 minute?

Distance is given by the integral of a velocity function. For this question, we will need to integrate over the interval of 0s to 60s.

$x(t) = \int_${0}^{60}$v(t),dt$

$x(60)=\int_${0}^{60}$(8t+4),dt = $(4(60)^2$$+4(60))-(4(0)^2$+4(0))$

$x(60)=4(3600)+240 = 14640\ m$

A particle traveling in a straight line accelerates uniformly from rest to $50:\up$\text{cm/s}$$ in $5:\up$\text{s}$$ and then continues at constant speed for an additional for an additional $3:\up$\text{s}$$ . What is the total distance traveled by the particle during the $8:\up$\text{s}$$ ?

A man runs with a velocity described by the function below.

What is the function for his acceleration?

The function for acceleration is the derivative of the function for velocity.

$a(t)= $\frac{dv(t)}{dt}$$

A man runs with a velocity as described by the function below.

How far does he travel in 1 minute?

Distance is given by the integral of a velocity function. For this question, we will need to integrate over the interval of 0s to 60s.

$x(t) = \int_${0}^{60}$v(t),dt$

$x(60)=\int_${0}^{60}$(8t+4),dt = $(4(60)^2$$+4(60))-(4(0)^2$+4(0))$

$x(60)=4(3600)+240 = 14640\ m$

A particle traveling in a straight line accelerates uniformly from rest to $50:\up$\text{cm/s}$$ in $5:\up$\text{s}$$ and then continues at constant speed for an additional for an additional $3:\up$\text{s}$$ . What is the total distance traveled by the particle during the $8:\up$\text{s}$$ ?

A man runs with a velocity described by the function below.

What is the function for his acceleration?

The function for acceleration is the derivative of the function for velocity.

$a(t)= $\frac{dv(t)}{dt}$$

A man runs with a velocity as described by the function below.

How far does he travel in 1 minute?

Distance is given by the integral of a velocity function. For this question, we will need to integrate over the interval of 0s to 60s.

$x(t) = \int_${0}^{60}$v(t),dt$

$x(60)=\int_${0}^{60}$(8t+4),dt = $(4(60)^2$$+4(60))-(4(0)^2$+4(0))$

$x(60)=4(3600)+240 = 14640\ m$