Question

A box accidentally drops from a truck traveling at a speed of 20.0 m/s and slides along the ground for a distance of 30.0 m Calculate a) The acceleration of the box (assuming it is constant while sliding)

Answer 1

a= - 6.667 m/s²

Step-by-step explanation:

Given that

The initial speed of the box ,u= 20 m/s

The final speed of the box ,v= 0 m/s

The distance cover by box ,s= 30 m

Lets take the acceleration of the box = a

We know that

v²= u ² + 2 a s

Now by putting the values in the above equation we get

0²=20² + 2 a x 30

`a=- (20^2)/(2* 30) \ m/s^2`

a= - 6.667 m/s²

Negative sign indicates that velocity and acceleration are in opposite direction.

Therefore the acceleration of the box will be - 6.667 m/s² .