Knowns:

\(\displaystyle v_0 = 20m/s\)

\(\displaystyle v_f=0m/s\)

\(\displaystyle a = -3.45m/s^2\)

\(\displaystyle t_0 = 0s\)

\(\displaystyle t_{reaction} = 0.360s\)

\(\displaystyle \Delta x = 19m\)

Unknowns:

\(\displaystyle \Delta x\) that Trevor actually travels = ?

Equation:

Since Trevor takes a moment to react before stepping on the breaks, determine how far Trevor travels during the reaction time. At this time he is traveling at a constant velocity.

\(\displaystyle v = \frac{\Delta x}{\Delta t}\)

\(\displaystyle \Delta x = v*\Delta t\)

\(\displaystyle \Delta x = 20m/s * 0.360s\)

\(\displaystyle \Delta x = 7.2m\)

Next, using kinematic equations to determine where Trevor stops his car based on the acceleration of his car.

\(\displaystyle v^f_2=v^0_2 +2a\Delta x\)

\(\displaystyle \Delta x =\frac{v^f_2-v^0_2}{2a}\)

\(\displaystyle \Delta x =\frac{0^2-(20m/s)^2}{2(-3.45m/s^2)}\)

\(\displaystyle \Delta x = 57.97m\) beyond the reaction point

Total distance traveled is the reaction time distance plus the distance traveled during the deceleration period.

total distance traveled \(\displaystyle = 7.2m +57.97m = 65.17m\)

To find the distance beyond the red light, subtract the distance traveled from the distance to the light

distance beyond red light \(\displaystyle = 65.17m-19m = 46.2m\)

Knowns:

\(\displaystyle v_{speeder} = 130km/h\)

\(\displaystyle v_{officer} =900km/h\)

\(\displaystyle t_{0 speeder} = 0 seconds\)

\(\displaystyle t_{0 officer}= 1 second\)

\(\displaystyle a=2.50m/s^2\)

Unknowns:

\(\displaystyle t_{final}=?\)

Equation:

To make things easier, set the frame of reference to the police officer.  This means that the police officer is traveling at 0km/h and the speeder is moving at a speed relative to the officer.

\(\displaystyle v_{speeder in reference to officer} = v_{speeder}- v_{officer}\)

\(\displaystyle v = 130km/h-90km/h\)

\(\displaystyle v= 40km/h\)

So the new relative velocities are:

\(\displaystyle v_{speeder} = 40km/h\)

\(\displaystyle v_{officer} =0km/h\)

This will help make the math easier.

Convert the relative velocity of the speeder to m/s.

\(\displaystyle \frac{40km}{h}*\frac{1000m}{1km}*\frac{1h}{60 min}*\frac{1min}{60s}= 11.11m/s\)

Now there are two equations that have to be true for the officer to catch up to the speeder.

\(\displaystyle \Delta x_{officer} =v_o\Delta t_{officer} + \frac{1}{2}a\Delta t^2_{officer}\)

\(\displaystyle v_{speeder}=\frac{\Delta x_{speeder}}{\Delta t_{speeder}}\)

There are two things that connect these equations.  First, the distance that the officer travels and the speeder travels must be the same.  Additionally the final velocity of both the officer and the speeder must be the same.

Rearrange the speeder equation for the distance traveled.

\(\displaystyle v_{speeder} * \Delta t_{speeder} = \Delta x_{speeder}\)

Set the two equations equal, as the distance traveled of the speeder is equal to the distance traveled by the officer.

\(\displaystyle v_o\Delta t_{officer} + \frac{1}{2}a\Delta t^2_{officer}=v_{speeder} * \Delta t_{speeder}\)

Remember that the reference point is at the officer’s perspective so the officer has an initial velocity of 0m/s.

\(\displaystyle 0m/s + \frac{1}{2}a\Delta t^2_{officer}=v_{speeder} * \Delta t_{speeder}\)

Rewrite the equation breaking up the change of time into time final and time initial.

\(\displaystyle \frac{1}{2}a(t_{f officer}- t_0 officer)^2=v_{speeder} * (t_{f speeder} -t_{0 speeder})\)

Distribute on the left side.

\(\displaystyle \frac{1}{2}a(t^2_{f officer}- t_{0 officer}t_{f officer} +t^2_{0 officer})=v_{speeder} * (t_{f speeder} -t_{0 speeder})\)

\(\displaystyle \frac{1}{2}at^2_{f officer}- \frac{1}{2}at_{0 officer}t_{f officer} +\frac{1}{2}at^2_{0 officer}=v_{speeder} * (t_{f speeder} -t_{0 speeder})\)

Substitute values

\(\displaystyle \frac{1}{2}(2.5m/s^2)t^2_{f officer}- \frac{1}{2}(2.5m/s^2)(1s)t_{f officer} +\frac{1}{2}(2.5m/s^2) (1s)^2=11.11m/s* (t_{f speeder} -0s)\)

Simplify

\(\displaystyle (1.25)t^2_{f officer}- 1.25t_{f officer} +1.25=11.11 t_{f speeder}\)

Remember that the final time for both the officer and the speeder are the same.  So rearrange this in the form of a quadratic equation.

\(\displaystyle (1.25)t^2_{f officer}- 12.36t_{f officer} +1.25=0\)

Use the quadratic formula to solve.  

Possible answers: \(\displaystyle t= 0.10s\) or \(\displaystyle t = 9.79s\)

Since the officer did not move until \(\displaystyle 1s\), then the second answer must be correct.

We are given final velocity and we know initial velocity is 0. We can use the following kinematics formula to relate final speed, initial speed, acceleration, and displacement:

\(\displaystyle v_f^2=v_0^2+2a\Delta x\)

The initial velocity is zero since Usain starts from rest. Therefore we can remove it from the equation.

\(\displaystyle v_f^2=2a\Delta x\)

Rearrange the equation to solve for acceleration.

\(\displaystyle \frac{v_f^2}{2\Delta x}=a\)

\(\displaystyle \frac{(13.4m/s)^2}{2(100m)}=a\)

\(\displaystyle a = .898m/s^2\)

Knowns:

\(\displaystyle v_0 = 0m/s\)

\(\displaystyle v_f =13m/s\)

\(\displaystyle \Delta t =31s\)

Unknowns:

\(\displaystyle a = ?\)

\(\displaystyle \Delta x = ?\)

Equation:

First we need to determine Peter’s acceleration. The best kinematic equation to use here is:

\(\displaystyle v_f=v_0+a\Delta t\)

\(\displaystyle \frac{v_f-v_0}{\Delta t}=a\)

\(\displaystyle \frac{13m/s-0m/s}{31s}=a\)

\(\displaystyle a = 0.42m/s^2\)

Once Peter’s acceleration is known it is possible to find the distance he traveled.

\(\displaystyle \Delta x = v_0 \Delta t+\frac{1}{2}a\Delta t^2\)

Peter’s initial velocity is \(\displaystyle 0m/s\) so this equation simplifies to

\(\displaystyle \Delta x =\frac{1}{2}a\Delta t^2\)

\(\displaystyle \Delta x =\frac{1}{2}(0.42m/s^2)(31)^2\)

\(\displaystyle \Delta x = 201.8m\)

\(\displaystyle \Delta x = v_0 \Delta t+\frac{1}{2}a\Delta t^2\)

The distance traveled is directly related to the square of the time traveled.  Therefore time if time is doubled, the value will be squared and therefore four times as great.  The distance traveled will be 4 times as far.

The can will have the same acceleration both up and down the hill since there is no friction.  Friction or other forces would cause a change in acceleration.  But since there is no friction, the can will travel up the hill, slow down and then accelerate back down the hill.

Knowns:

\(\displaystyle \Delta x_{total} = 10,000m\)

\(\displaystyle \Delta t_{total} = 30 min\)

\(\displaystyle \Delta t_{constant speed} = 26 min\)

\(\displaystyle \Delta x_{accelerating} = 1500m\)

\(\displaystyle a = 0.30m/s^2\)

Unknown:

\(\displaystyle \Delta t_{accelerating}= ?\)

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

\(\displaystyle v = \frac{\Delta x}{\Delta t}\)

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

\(\displaystyle 26 min *\frac{60s}{1 min}= 1560s\)

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

\(\displaystyle 10,000m-1,500m = 8,500m\)

Use these values to determine the velocity of the runner.

\(\displaystyle v = \frac{8,500m}{1560s}\)

\(\displaystyle v = 5.45m/s\)

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

\(\displaystyle 30min - 26min = 4 min\) left in race

\(\displaystyle 4min *\frac{60s}{1 min} = 240s\)

\(\displaystyle v = \frac{1500m}{240s}\)

\(\displaystyle v = 6.25 m/s\)

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

\(\displaystyle v_f= v_0+a\Delta t\)

Rearrange for time

\(\displaystyle \Delta t = \frac{v_f-v_0}{a}\)

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

\(\displaystyle \Delta t = 2.67s\)

Explanation:

We are given the initial velocity, time, and final velocity (zero because the ball stops). Using these values and the appropriate motion equation, we can solve for the acceleration.

Acceleration is given by the change in velocity over time:

\(\displaystyle a = \frac{v_f-v_0}{\Delta t}\)

We can use our values to solve for the acceleration.

\(\displaystyle a = \frac{0m/s - 8.1m/s}{4.2s}\)

\(\displaystyle a = -1.93m/s^2\)

\(\displaystyle v_f = v_0+a\Delta t\)

There is a direct relationship between the final velocity and the time traveled.  Therefore if the time is doubled, the velocity would double as well.