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A truck starts from rest and accelerates uniformly at 2.0 ms-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.)
A truck starts from rest and accelerates uniformly at 2.0 ms-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.)
CBSE | Class 11 | Laws of Motion | NCERT | Physics
A truck starts from rest and accelerates uniformly at 2.0 ms-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.)

u = 0 is the initial velocity.

a = 2 ms-2, a = 2 ms-2, a = 2 ms-2, a = 2 m

 

t=10s t=10s t=10s t=10

 

We get v = u + at using the equation v = u + at.

 

20 m/s = v = 0 + 2 x 10

 

v = 20 m/s is the final velocity.

 

In the absence of air resistance, the horizontal component of velocity stays constant at time t = 11 sec.

 

20 m/s Vx

 

The equation gives the vertical component of velocity.

 

Vy = ayt + u

 

Here, t = 11 – 10 = 1s and ay= a = 10 m/s are the values.

 

As a result, the stone’s resulting velocity, v, is

v = ( vx2 + vy2)½

v = (20 2 + 10 2)1/2

v = 22.36 m/s

tan θ = vy/vx = 10/20 = ½ = 0.5

 

from the horizontal = tan -1 (0.5 ) = 26. 560

 

(b) The horizontal force on the stone is zero when it is dropped from the truck. The stone’s only acceleration is that caused by gravity, which is 10 m/s2 and works vertically downwards.

i

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