Answer:
Acceleration of the car will be equal to [tex]a=1.805m/sec^2[/tex]
Explanation:
As the car is initially stop so initial velocity of the car u = 0 m /sec
It is given that after moving a distance of 101 m velocity of car is 19.1 m/sec
So final velocity of the car v=19.1 m/sec
We have to find the acceleration of the car
From third equation of motion we know that [tex]v^2=u^2+2as[/tex]
So [tex]19.1^2=0^2+2\times a\times 101[/tex]
[tex]a=1.805m/sec^2[/tex]
So acceleration of the car will be equal to [tex]a=1.805m/sec^2[/tex]