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 [tex]a_1=5.9m/sec^2[/tex]
Acceleration of the stock car [tex]a_2=3.6m/sec^2[/tex]
When 1st car overtakes the second car then distance traveled by both the car will be same
(a) So [tex]s_1=s_2[/tex]
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 [tex]\frac{1}{2}\times 5.9\times t^2=\frac{1}{2}\times (t+1)^2\times 3.6[/tex]
After solving t = 3.56 sec
(b) From second equation of motion [tex]s=ut+\frac{1}{2}at^2=0\times 3.56+\frac{1}{2}\times 5.9\times 3.56^2=37.38712m[/tex]
(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
