The question seems to be expecting a comparative answer.
Let us consider Scenario 1 first.
We have the variables listed as follows:
Initial Velocity [tex]V_{i}[/tex] = v m/s
Final Velocity [tex]V_{f}[/tex] = 0, since the car is stopping.
Acceleration = - a [tex]m/s^{2}[/tex] , 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 [tex]V_{f}^{2} } = V_{i}^{2} } + 2aS[/tex]
Plugging in what we know, we get [tex]0 = v^{2} - 2aS[/tex]
Hence, stopping distance in Scenario 1 is [tex]S = \frac{v^{2} }{2a}[/tex]
Now, considering Scenario 2, we have
Initial Velocity [tex]V_{i}[/tex] = 5v m/s
Final Velocity [tex]V_{f}[/tex] = 0
Acceleration = - a [tex]m/s^{2}[/tex], since it is given to be same in both the cases.
Let the stopping distance this time be [tex]S_{2}[/tex]
We can use the same equation for this scenario too, i.e. [tex]V_{f}^{2} } = V_{i}^{2} } + 2aS[/tex]
Plugging in what we have, we get [tex]0 = (5v)^{2} - 2aS_{2}[/tex]
Solving this for [tex]S_{2}[/tex], we get [tex]S_{2} = 25. \frac{v^{2} }{2a}[/tex]
This can be written in terms of S as [tex]S_{2} = 25.S[/tex]
Thus, the car's new stopping distance will be 25 times compared to the older one!
