Linear Motion

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Questions 1 - 10

An athlete kicks a ball into the air. It travels $\small 42m$ and is in the air for $\small 4.5s$ . How fast must the athlete run from the point where he kicks the ball in order to catch it before it lands?

$9.33\frac{m}{s}$

$10.21\frac{m}{s}$

$9.19\frac{m}{s}$

$18.7\frac{m}{s}$

$8.74\frac{m}{s}$

Explanation

To solve this problem, we need to understand what speed is. Speed is the distance covered in a given amount of time.

$\text{speed}=\frac{\text{distance}}{\text{time}}$

In our problem, we need to find the speed of the athlete. We are given the distance the athlete must cover, which is equal to the distance traveled by the ball.

We are also told how much time the athlete has to cover the distance, which is equal to the time the ball is in the air.

Use these values and the equation for speed to find the speed that the athlete must run.

$s=\frac{d}{t}=\frac{42m}{4.5s}$

$s=9.33\frac{m}{s}$

A runner wants to complete a run in less than . After running at constant speed for exactly , the runner still has left to run. The runner must then accelerate at for how many seconds in order to reach a final velocity that will allow them to complete the left of the race in the desired time?

Explanation

Knowns:

$\Delta x_{total} = 10,000m$

$\Delta t_{total} = 30 min$

$\Delta t_{constant speed} = 26 min$

$\Delta x_{accelerating} = 1500m$

Unknown:

$\Delta t_{accelerating}= ?$

Equation:

The first thing is to determine the initial velocity of the runner before the runner accelerates for the final portion of the race. Since the runner is traveling at a constant velocity

$v = \frac{\Delta x}{\Delta t}$

Next, convert the time during the constant speed portion to seconds.

$26 min *\frac{60s}{1 min}= 1560s$

Determine the amount of distance traveled while at a constant speed.

Use these values to determine the velocity of the runner.

$v = \frac{8,500m}{1560s}$

Next determine the final velocity needed for the runner to finish out the race in the remaining time.

left in race

$4min *\frac{60s}{1 min} = 240s$

$v = \frac{1500m}{240s}$

Finally use the kinematic equations to calculate the time needed to get to this velocity.

$v_f= v_0+a\Delta t$

Rearrange for time

$\Delta t = \frac{v_f-v_0}{a}$

$\Delta t = \frac{6.25m/s-5.45m/s}{0.30m/s^2}$

$\Delta t = 2.67s$

While traveling along a highway, a specific automobile is capable of an acceleration of about $1.6m/s^{2}$ . At this rate, how long does it take to accelerate from to ?

Explanation

Knowns:

$a = 1.6m/s^2{}$

$v_{0}= 65km/h$

$v_{f}= 120km/h$

Unknowns:

$\Delta t = ?$

Equation:

$a=\frac{{}v_{f} - v_{0}}{\Delta t}$

The most important thing to recognize here is that the initial and final velocities are not in the correct units. The first step is to convert both of these values to .

$\frac{65km}{h} * \frac{1h}{60 min} * \frac{1 min}{60 s} *\frac{100m}{km} = 18.06m/s$

$\frac{120km}{h} * \frac{1h}{60 min} * \frac{1 min}{60 s} *\frac{100m}{km} = 33.33m/s$

Then rearrange your equation to solve for (the missing variable).

${\Delta t}=\frac{{}v_{f} - v_{0}}{a}$

Now plug in the variables and solve.

${\Delta t}=\frac {33.33m/s - 18.06m/s}{1.6m/s^2}$

$\Delta t = 9.55s$

$9.33\frac{m}{s}$

$10.21\frac{m}{s}$

$9.19\frac{m}{s}$

$18.7\frac{m}{s}$

$8.74\frac{m}{s}$

Explanation

To solve this problem, we need to understand what speed is. Speed is the distance covered in a given amount of time.

$\text{speed}=\frac{\text{distance}}{\text{time}}$

In our problem, we need to find the speed of the athlete. We are given the distance the athlete must cover, which is equal to the distance traveled by the ball.

We are also told how much time the athlete has to cover the distance, which is equal to the time the ball is in the air.

Use these values and the equation for speed to find the speed that the athlete must run.

$s=\frac{d}{t}=\frac{42m}{4.5s}$

$s=9.33\frac{m}{s}$

Leslie rolls a ball out of a window from above the ground, such that the initial y-velocity is zero. How long will it be before the ball hits the ground?

Explanation

We are given the initial velocity, acceleration, and distance traveled. Using the equation below, we can solve for the time. Remember that the initial velocity is 0m/s so it drops out of the equation.

$\Delta x = v_0 \Delta t+\frac{1}{2}a\Delta t^2$

$\Delta x =\frac{1}{2}a\Delta t^2$

$\Delta t = \sqrt{\frac{2\Delta x}{a}}$

$\Delta t = \sqrt{\frac{2(0m-10m)}{(-9.8m/s^2)}}$

The distance is negative, which makes since because the ball is traveling downward. Also when taking the square root, only the positive value is needed as it is impossible to have negative time.

$\Delta t = 1.43s$

Suppose a recreational biker averages $8.05\frac{m}{s}$ on a twenty-mile ride, equal to . A professional biker has an average speed of $11.1\frac{m}{s}$ . The professional happens to be riding on the same path, but started behind the recreational biker. The two are both headed for the same destination. Who would reach the end of the path first, and how far behind would the other biker be?

The professional finishes first and the other biker is behind

Both bikers reach destination at same time

The recreational finishes first and the other biker is behind

Explanation

To solve this problem we can simply examine each biker separately and see how long it would take them to reach the destination. Let's begin with the recreational biker. The recreational biker needs to ride at $8.05\frac{m}{s}$ . Using the definition of velocity, we can find his final time.

$v=\frac{d}{t}$

$t=\frac{d}{v}$

$t=\frac{3.22*10^4m}{8.05\frac{m}{s}}$

The professional has a distance of , plus the that he's behind.

We know that the velocity of the professional biker is $11.1\frac{m}{s}$ . Using this velocity and his total distance, we can find the time that it takes him to reach the end of the path.

$v=\frac{d}{t}$

$t=\frac{d}{v}$

$t=\frac{4.42*10^4m}{11.1\frac{m}{s}}$

The time of the professional biker is less than that of the recreational biker, meaning that the professional will finish first. Now we need to find the distance between the two bikers at this point. Use the recreational biker's velocity and the time difference between the two bikers to solve for the distance that the recreational biker has left on the path.

The recreational biker will ride for at $8.05\frac{m}{s}$ to finish the path.

$v=\frac{d}{t}$

$d=(8.05\frac{m}{s})(18s)$

$d=144.9m\approx145m$

Laura throws a ball vertically. She notices it reaches a maximum height of 10 meters. What was the initial velocity of the ball?

$\small g=-9.8\frac{m}{s^2}$

$14\frac{m}{s}$

$196\frac{m}{s}$

$144\frac{m}{s}$

$29\frac{m}{s}$

$22.2\frac{m}{s}$

Explanation

Remember that at the highest point, the velocity in the y-direction is equal to zero. Using the given values and the equation below, we can solve for the initial velocity.

$V_f^2=V_i^2+2a\Delta y$

$(0\frac{m}{s})^2=V_i^2+2(-9.8\frac{m}{s^2})(10m)$

$0\frac{m^2}{s^2}=V_i^2-196\frac{m^2}{s^2}$

$196\frac{m^2}{s^2}=V_i^2$

$\sqrt{196\frac{m^2}{s^2}}=\sqrt{V_i^2}$

$14\frac{m}{s}=V_i$

A man stands on a tall ladder of height . He leans over a little too far and falls off the ladder. What would be the best way to describe his fall?

Parabolic motion

One-dimensional motion

Circular motion

We would need to know his mass in order to determine the type of motion

We would need to know air resistance in order to determine his type of motion

Explanation

The man's fall will be parabolic as there will be both horizontal and vertical components. His vertical component of the fall will be standard free-fall caused by his acceleration due to gravity. His horizontal component of the fall will come from him "leaning too far" in one direction. Even a small horizontal velocity will create a horizontal trajectory.

This is why when people lean and fall off of ladders they either try to grab onto the ladder (try to negate their horizontal velocity) or fall a small distance away from the base of the ladder.

Sam throws a rock off the edge of a tall building at an angle of $35^{\circ}$ from the horizontal. The rock has an initial speed of $30\frac{m}{s}$ .

$\small g=-9.8\frac{m}{s^2}$

At what height above the ground will the rock change direction?

Explanation

Even though the problem gives us an initial velocity, we need to break it down into horizontal and vertical components.

$v_{i} *\sin(\Theta)=v_{iy}$

We can plug in the given values and find the vertical velocity.

$30\frac{m}{s}* \sin(35^{\circ})=v_{iy}$

$30\frac{m}{s}* (0.57)=v_{iy}$

$17.21\frac{m}{s}=v_{iy}$

Remember that the vertical velocity at the highest point of a parabola is zero. Now that we know the initial and final vertical velocities, we can plug our values into an equation to solve for the maximum height.

$v_{f}^2=v_i^2+2a\Delta y$

$(0\frac{m}{s})^2=(17.21\frac{m}{s})^2+2(-9.8\frac{m}{s^2})\Delta y$

$(0\frac{m}{s})=(296.09\frac{m}{s})+(-19.6\frac{m}{s^2})\Delta y$

$-(296.09\frac{m}{s})=(-19.6\frac{m}{s^2})\Delta y$

$\frac{-296.09\frac{m}{s}}{-19.6\frac{m}{s^2}}=\Delta y$

$15.11m=\Delta y$

Remember, $\Delta y$ only tells us the CHANGE in the vertical direction. The rock started at the top of a tall building, then rose an extra .

Its highest point is above the ground.

Laurence throws a rock off the edge of a tall building at an angle of $20^{\circ}$ from the horizontal with an initial speed of $31\frac{m}{s}$ .

$g=-9.8\frac{m}{s^2}$ .

How long is the rock in the air?

Explanation

The given velocity won't help us much here. We need to break it down into horizontal and vertical components.

Use the sine function with the initial velocity and angle to find the vertical velocity.

$v_{i}*\sin(\theta)=v_{iy}$

$31\frac{m}{s}*\sin(20^{\circ})=v_{iy}$

$31\frac{m}{s}* (0.34)=v_{iy}$

$10.6\frac{m}{s}=v_{iy}$

Now we need to work on finding the time in the air. To do this, we need to break the rock's path into two parts. The first part is the time that the rock is rising to its maximum height, and the second part is the time that it is falling from the highest point to the ground.

For the first part, we can assume that the final vertical velocity is zero, since this will be the top of the parabola. Using the initial velocity, final velocity, and gravity, we can solve for the time to travel this portion of the path.

$0\frac{m}{s}=10.6\frac{m}{s}+(-9.8\frac{m}{s^2})t_1$

$-10.6\frac{m}{s}=-9.8\frac{m}{s^2}\cdot t_1$

$\frac{-10.6\frac{m}{s}}{-9.8\frac{m}{s^2}}=t_1$

This is the time that the rock is traveling upward. Now we need to focus on the time that the rock travels downward. We don't know the final velocity at the end of the parabola, so we can't use the equation from the first part. If we can find the total height of the parabola, we can use a different equation to solve for time.

$\Delta y =v_it+\frac{1}{2}at^2$

Remember that the vertical velocity at the highest point of a parabola is zero. Now that we know the initial vertical velocity, we can plug this into an equation to solve for the distance that the rock travels from its maximum height to the original height.

$v_f^2=v_i^2+2a\Delta y$

$(0\frac{m}{s})^2=(10.6\frac{m}{s})^2+2(-9.8\frac{m}{s^2})\Delta y$

$(0\frac{m}{s})=(112.36\frac{m}{s})+(-19.6\frac{m}{s^2})\Delta y$

$-(112.36\frac{m}{s})=(-19.6\frac{m}{s^2})\Delta y$

$\frac{-112.36\frac{m}{s}}{-19.6\frac{m}{s^2}}=\Delta y$

$5.73m=\Delta y$

Remember, $\Delta y$ only tells us the CHANGE in height. Since the rock started at the top of a tall building, if it rose an extra , then at its highest point it is above the ground.

This means that our $\Delta y$ for the second equation will be (the change in height is negative because it travels downward.) Use this total distance and the velocity at the top of the peak (zero) to solve for the time that the rock travels down.