Answer:
(a) Distance traveled = 75.3846 m
(b) Velocity of car at that instant will be 14 m/sec
Explanation:
We have given acceleration of the car [tex]a=1.3m/sec^2[/tex]
Initial velocity of the cart u = 0 m/sec
(a) According to second equation of motion we know that [tex]s=ut+\frac{1}{2}at^2[/tex]
So distance traveled by car [tex]s_c=0\times t+\frac{1}{2}\times 1.3t^2=0.65t^2[/tex]
As the truck is moving with constant speed
So distance traveled by truck [tex]s_t=ut=7t[/tex]
As the truck overtakes the car
So [tex]s_c=s_t[/tex]
[tex]0.65t^2=7t[/tex]
[tex]t=10.769sec[/tex]
So distance traveled [tex]s_c=s_t=7\times 10.769=75.3846m[/tex]
(b) From second equation of motion we know that v = u+at
So v = 0+1.3×10.769 = 14 m /sec