Answer:
The ratio of stopping distances is 4 i.e by a factor 4 the stopping distances differ
Explanation:
Using 3rd equation of motion we have
For car 1
[tex]v_{1}^{^{2}}=u_{1}^{2}+2a_{1}s_{1}[/tex]
For car 2 [tex]v_{2}^{^{2}}=u_{2}^{2}+2a_{2}s_{2}[/tex]
Since the initial speed of both the cars are equal and when the cars stop the final velocities of both the cars become zero thus the above equations reduce to
[tex]u^{2}=-2a_{1}s_{1}\\\\s_{1}=\frac{-u^{2}}{2a_{1}}[/tex].............(i)
Similarly for car 2 we have
[tex]s_{2}=\frac{-u^{2}}{2a_{2}}[/tex]..................(ii)\
Taking ratio of i and ii we get
[tex]\frac{s_{1}}{s_{2}}=\frac{a_{2}}{a_{1}}[/tex]
Let
[tex]\frac{a_{2}}{a_{1}}=4[/tex]
Thus
[tex]\frac{s_{1}}{s_{2}}=4[/tex]
The ratio of stopping distances is 4
