Answer:
(a) Time will be t = 3.56 sec
(b) Distance traveled by car when they are side by side is 37.38712 m
(b) Velocity of race car = 21.004 m/sec
velocity of stock car = 12.816 m/sec
Explanation:
We have given acceleration of the car ![a_1=5.9m/sec^2](https://tex.z-dn.net/?f=a_1%3D5.9m%2Fsec%5E2)
Acceleration of the stock car ![a_2=3.6m/sec^2](https://tex.z-dn.net/?f=a_2%3D3.6m%2Fsec%5E2)
When 1st car overtakes the second car then distance traveled by both the car will be same
(a) So ![s_1=s_2](https://tex.z-dn.net/?f=s_1%3Ds_2)
As both car starts from rest so initial velocity of both car will be 0 m/sec
It is given that stock car leaves 1 sec before
So ![\frac{1}{2}\times 5.9\times t^2=\frac{1}{2}\times (t+1)^2\times 3.6](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Ctimes%205.9%5Ctimes%20t%5E2%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%28t%2B1%29%5E2%5Ctimes%203.6)
After solving t = 3.56 sec
(b) From second equation of motion ![s=ut+\frac{1}{2}at^2=0\times 3.56+\frac{1}{2}\times 5.9\times 3.56^2=37.38712m](https://tex.z-dn.net/?f=s%3Dut%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2%3D0%5Ctimes%203.56%2B%5Cfrac%7B1%7D%7B2%7D%5Ctimes%205.9%5Ctimes%203.56%5E2%3D37.38712m)
(c) From first equation pf motion v = u+at
So velocity of race car v = 0+5.9×3.56 = 21.004 m/sec
Velocity of stock car v = 0+ 3.6×3.56 = 12.816 m/sec