Question

What is the stopping distance if the car is initially traveling at speed 5.0v? assume that the acceleration due to the braking is the same in both cases?

Answer 1

The question seems to be expecting a comparative answer.

Let us consider Scenario 1 first.

We have the variables listed as follows:

Initial Velocity  $V_{i}$ = v m/s

Final Velocity  $V_{f}$ = 0, since the car is stopping.

Acceleration  = - a  $m/s^{2}$ , since the car is slowing down.

Let the stopping distance in this case be S

We can arrive at the expression for S by using the equation $V_{f}^{2} } = V_{i}^{2} } + 2aS$

Plugging in what we know, we get $0 = v^{2} - 2aS$

Hence, stopping distance in Scenario 1 is $S = \frac{v^{2} }{2a}$

Now, considering Scenario 2, we have

Initial Velocity $V_{i}$ = 5v m/s

Final Velocity  $V_{f}$ = 0

Acceleration  = - a  $m/s^{2}$, since it is given to be same in both the cases.

Let the stopping distance this time be $S_{2}$

We can use the same equation for this scenario too, i.e. $V_{f}^{2} } = V_{i}^{2} } + 2aS$

Plugging in what we have, we get $0 = (5v)^{2} - 2aS_{2}$

Solving this for  $S_{2}$, we get $S_{2} = 25. \frac{v^{2} }{2a}$

This can be written in terms of S as $S_{2} = 25.S$

Thus, the car's new stopping distance will be 25 times compared to the older one!