A truck starts from rest and accelerates uniformly at 2.0 m s¯². At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What are the (a) velocity, and (b) acceleration of the stone at t = 11s ? (Neglect air resistance.)

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Solution

Given: The initial speed of the truck is 0 m/s , the acceleration of the truck is 2.0 m/ s 2 , the height of the top of the truck is 6 m, and the time interval at which the stone is thrown from the top of the truck is 10 s.

(a)

According to first equation of motion, the final velocity of the truck is,

v=u+at(1)

By substituting the given values in the above equation, we get

v=0 m/s +( 2.0 m/ s 2 )( 10 s ) =20 m/s

According to Newton’s first law, the horizontal velocity of the stone remains constant throughout the motion of the stone, that is 20 m/s.

At t=11 s,

The horizontal velocity is,

u x =20 m/s

The vertical velocity after the stone is thrown is,

v y =u+gt

By substituting the given values in the above equation, we get

v y =0 m/s +( 10 m/ s 2 )( 1 s ) =10 m/s

The vertical and horizontal components of velocity are shown below:

The resultant velocity of the stone is given as,

v r = v x 2 + v y 2 = ( 20 m/s ) 2 + ( 10 m/s ) 2 =22.4 m/s

The tangent of the angle with the horizontal is given as,

tanθ= v y v x θ= tan −1 ( 10 20 ) = tan −1 ( 1 2 )

Thus, the velocity of the stone at t=11 s is 22.4 m/s at an angle tan −1 ( 1 2 ) with the horizontal.

(b)

The horizontal acceleration of the stone is zero because it moves with constant velocity as the horizontal force becomes zero when it is dropped from the truck.

The stone moves under gravitational force. Therefore, the net acceleration of the stone is 10 m/ s 2 vertically downward.

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