Question

X0= -4, V0=3, A0=-0.520. If instead v0 = 2.27 in which location now does the car charges direction?

Answer 1

The initial positon of the car is X0=-4( assuming that the car is on the negativer x axis)

velocity is V0 = +3 , hence we can say it is moving towards the positive x direction

however the acceleration A0 = -0.520 , which means the body is decelerating , so the final veloctiy will become 0 at some point and will change direction . we can use the kinematical equations to get the distance from its initial postion where the body will aquire zero velocity and change direction.

( S stands for distance travelled by the car, note that final velocity V should be taken as zero)

Solving , gives S =8.65. BUT , since the starting point of the car on the axis is -4 . The point on the x- axis where the car will change direction is 8.65-4 = +4.65

Now it is said the velocity is V0= 2.27

Using the same logic as before and the kinematical equations

we get

on solving S = 4.954

BUT, since the starting point on the x axis isx=-4

the point on the x axis where the car turns is 4.95 - 4 = +0.95