Connecting Position, Velocity, and Acceleration
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AP Calculus BC › Connecting Position, Velocity, and Acceleration
A particle has velocity $v(t)=\sin t+t$ for $0\le t\le \pi$ and $s(0)=-1$; find $s(\pi)$.
$s(\pi)=-1+\int_0^{\pi}(\sin t),dt$
$s(\pi)=-1+\int_0^{\pi}(\cos t+1),dt$
$s(\pi)=-1+(\sin \pi+\pi)$
$s(\pi)=-1+\int_0^{\pi}(\sin t+t),dt$
$s(\pi)=\int_0^{\pi}(\sin t+t),dt$
Explanation
This problem involves using integration in kinematics to connect velocity to position. To find the position at t=π, integrate the velocity function v(t)=sin t + t from 0 to π, which gives the change in position over that interval. Then, add the initial position s(0)=-1 to account for the starting point. This yields s(π)=-1 + ∫ from 0 to π of (sin t + t) dt, as the integral computes the net displacement. A tempting distractor is choice C, which evaluates the velocity at t=π and adds it to the initial position, mistaking velocity for displacement. In general, when solving initial value problems in kinematics, determine the constant of integration by applying the initial condition after finding the antiderivative.
A drone’s acceleration is $a(t)=\frac{6}{(t+1)^2}$ with $v(2)=4$; what is $v(5)$?
$v(5)=4+\int_2^5\frac{6}{t+1},dt$
$v(5)=4+\int_2^5\frac{6}{(t+1)^2},dt$
$v(5)=4+\frac{6}{(5+1)^2}$
$v(5)=\int_2^5\frac{6}{(t+1)^2},dt$
$v(5)=4+\int_0^5\frac{6}{(t+1)^2},dt$
Explanation
This problem involves using integration in kinematics to connect acceleration to velocity. To find the velocity at t=5, integrate the acceleration function a(t)=6/(t+1)² from 2 to 5, which gives the change in velocity over that interval. Then, add the initial velocity v(2)=4 to obtain the final velocity. This results in v(5)=4 + ∫ from 2 to 5 of 6/(t+1)² dt, ensuring the limits align with the initial condition. A tempting distractor is choice D, which adds the acceleration at t=5 to the initial velocity, confusing instantaneous values with integrated change. In general, when solving initial value problems in kinematics, determine the constant of integration by applying the initial condition after finding the antiderivative.
A car’s acceleration is $a(t)=4t-6$ (m/s$^2$) with $v(0)=5$; what is $v(2)$?
$v(2)=5+(4\cdot2-6)$
$v(2)=5+\int_0^2(4-6t),dt$
$v(2)=5+\int_0^2(4t-6)',dt$
$v(2)=\int_0^2(4t-6),dt$
$v(2)=5+\int_0^2(4t-6),dt$
Explanation
This problem involves using integration in kinematics to connect acceleration to velocity. To find the velocity at t=2, integrate the acceleration function a(t)=4t-6 from 0 to 2, which gives the change in velocity over that interval. Then, add the initial velocity v(0)=5 to obtain the final velocity. This results in v(2)=5 + ∫ from 0 to 2 of (4t-6) dt, where the integral represents the net change due to acceleration. A tempting distractor is choice C, which adds the acceleration at t=2 to the initial velocity, confusing instantaneous acceleration with the accumulated change. In general, when solving initial value problems in kinematics, determine the constant of integration by applying the initial condition after finding the antiderivative.
A particle’s velocity is $v(t)=\ln(t+1)$ for $0\le t\le 3$ and $s(1)=5$; find $s(3)$.
$s(3)=5+\ln(4)$
$s(3)=\int_1^3\ln(t+1),dt$
$s(3)=5+\int_1^3\ln(t+1),dt$
$s(3)=5+\int_1^3\frac{1}{t+1},dt$
$s(3)=5+\int_0^3\ln(t+1),dt$
Explanation
This problem involves using integration in kinematics to connect velocity to position. To find the position at t=3, integrate the velocity function v(t)=ln(t+1) from 1 to 3, which gives the change in position over that interval. Then, add the initial position s(1)=5 to account for the starting point. This yields s(3)=5 + ∫ from 1 to 3 of ln(t+1) dt, with limits matching the given initial time. A tempting distractor is choice D, which adds the velocity at t=3 to the initial position, mistaking instantaneous velocity for displacement. In general, when solving initial value problems in kinematics, determine the constant of integration by applying the initial condition after finding the antiderivative.
A particle moves with velocity $v(t)=t^3-6t$ for $0 \le t \le 2$ and $s(0)=0$; find $s(2)$.
$s(2)=\int_0^2(3t^2-6),dt$
$s(2)=\int_0^2(t^3-6t),dt$
$s(2)=\int_0^2(t^3),dt$
$s(2)=t^3-6t\big|_{t=2}$
$s(2)=0+\int_0^2(t^3-6t),dt$
Explanation
This problem involves using integration in kinematics to connect velocity to position. To find the position at t=2, integrate the velocity function $v(t)=t^3-6t$ from 0 to 2, which gives the change in position over that interval. Then, add the initial position s(0)=0 to account for the starting point. This yields $s(2)=0 + \int_0^2 (t^3-6t) , dt$, as the integral computes the net displacement. A tempting distractor is choice C, which evaluates the velocity at t=2 without integrating, ignoring the accumulation over time. In general, when solving initial value problems in kinematics, determine the constant of integration by applying the initial condition after finding the antiderivative.
A particle has velocity $v(t)=3t^2-4t+1$ for $0\le t\le 3$ and position $s(0)=2$; find $s(3)$.
$s(3)=\int_0^3(3t^2-4t+1),dt$
$s(3)=2+(3(3)^2-4(3)+1)$
$s(3)=2+\int_0^3(3t^2-4t),dt$
$s(3)=2+\int_0^3(6t-4),dt$
$s(3)=2+\int_0^3(3t^2-4t+1),dt$
Explanation
This problem involves using integration in kinematics to connect velocity to position. To find the position at t=3, integrate the velocity function v(t)=3t²-4t+1 from 0 to 3, which gives the change in position over that interval. Then, add the initial position s(0)=2 to account for the starting point. This yields the expression s(3)=2 + ∫ from 0 to 3 of (3t²-4t+1) dt, as the integral computes the net displacement. A tempting distractor is choice C, which incorrectly evaluates the velocity function at t=3 and adds it to the initial position, mistaking instantaneous velocity for displacement. In general, when solving initial value problems in kinematics, determine the constant of integration by applying the initial condition after finding the antiderivative.
A particle has acceleration $a(t)=4\cos(2t)$ for $0\le t\le\frac{\pi}{2}$ with $v(0)=-1$; what is $v!\left(\frac{\pi}{2}\right)$?
-5
-3
-1
1
3
Explanation
This problem involves kinematics integration to find velocity from acceleration. Since acceleration a(t) = 4cos(2t) is the derivative of velocity, we integrate using the chain rule in reverse: v(t) = ∫4cos(2t)dt = 4·(1/2)sin(2t) + C = 2sin(2t) + C. Using the initial condition v(0) = -1, we find 2sin(0) + C = -1, so 0 + C = -1, giving C = -1. Therefore v(t) = 2sin(2t) - 1. Evaluating at t = π/2 gives v(π/2) = 2sin(π) - 1 = 2(0) - 1 = -1. A common error is forgetting to divide by the coefficient of t when integrating, which would incorrectly give 4sin(2t). Always account for the chain rule when integrating composite functions.
A motorcycle’s acceleration is $a(t)=2\cos t$ with $v(0)=-1$; what is $v\left(\frac{\pi}{2}\right)$?
$v\left(\frac{\pi}{2}\right)=\int_0^{\pi/2}2\cos t,dt$
$v\left(\frac{\pi}{2}\right)=-1+2\cos\left(\frac{\pi}{2}\right)$
$v\left(\frac{\pi}{2}\right)=-1+\int_0^{\pi/2}2\cos t,dt$
$v\left(\frac{\pi}{2}\right)=-1+\int_0^{\pi/2}\cos t,dt$
$v\left(\frac{\pi}{2}\right)=-1+\int_0^{\pi/2}(-2\sin t),dt$
Explanation
This problem involves using integration in kinematics to connect acceleration to velocity. To find the velocity at t=π/2, integrate the acceleration function a(t)=2 cos t from 0 to π/2, which gives the change in velocity over that interval. Then, add the initial velocity v(0)=-1 to obtain the final velocity. This results in v(π/2)=-1 + ∫ from 0 to π/2 of 2 cos t dt, where the integral represents the net change due to acceleration. A tempting distractor is choice C, which adds the acceleration at t=π/2 to the initial velocity, mistaking instantaneous acceleration for the total change. In general, when solving initial value problems in kinematics, determine the constant of integration by applying the initial condition after finding the antiderivative.
An object’s velocity is $v(t)=\frac{2}{1+t^2}$ (m/s) and $s(0)=-3$ m; find $s(1)$.
$\int_{0}^{1}\frac{2}{1+t^2},dt$
$-3+\int_{0}^{1}\frac{2}{1+t^2},dt$
$-3+\left.\frac{2}{1+t^2}\right|_{0}^{1}$
$-3+\int_{0}^{1}\frac{-4t}{(1+t^2)^2},dt$
$-3+v(1)$
Explanation
This question requires kinematics integration to find position from velocity. Using $s(t) = s(0) + \int_0^t v(u), du$, with $s(0) = -3$ and $v(t) = \frac{2}{1+t^2}$, we get $s(1) = -3 + \int_0^1 \frac{2}{1+t^2}, dt$. Choice E incorrectly shows the derivative of the velocity function rather than integrating the velocity itself. When finding position from velocity, integrate the velocity function and add the initial position value.
A particle has velocity $v(t)=t\sqrt{t}$ (m/s) and $s(1)=2$ m; determine $s(4)$.
$2+v(4)$
$\int_{1}^{4}t\sqrt{t},dt$
$2+\int_{1}^{4}t\sqrt{t},dt$
$2+\left.t\sqrt{t}\right|_{1}^{4}$
$2+\int_{1}^{4}\frac{3}{2}\sqrt{t},dt$
Explanation
This question tests kinematics integration for position from velocity. Using $s(t) = s(t_0) + \int_{t_0}^t v(u), du$, with $s(1) = 2$ and $v(t) = t\sqrt{t} = t^{3/2}$, we get $s(4) = 2 + \int_1^4 t^{3/2}, dt$. Choice E incorrectly shows $(3/2)\sqrt{t}$, which appears to be a confused attempt at the antiderivative rather than the velocity function to integrate. To find position change, integrate the velocity function over the time interval and add the initial position.