Question

Question 5.11 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 = 11 s? (Neglect air resistance.)

Chapter Laws Of Motion Page 110

Answer 1

Truck starts from rest so, initial velocity = 0 .
acceleration = 2m/s²
Time = 10sec


(A) Velocity of truck at t = 10sec
V= u + at
= 0 + 2× 10 = 20m/s
Hence , horizontal velocity of stone = 20m/s

Vertical velocity :
When stone is dropped from the top of truck , velocity of it after 1 sec { 10sec to 11 sec }
V = U + at
initial velocity of stone in vertical direction = 0
V = 0 -9.8× 1 = -9.8 m/sec
Hence, in vector quantity ,velocity of stone is
V = 20i -9.8j
magnitude of it = √{20²+(9.8)²}
= 22.4 m/s
Let ∅makes with horizontal
Tan∅ = Vy/Vx = 9.8/20 =0.49
∅ = tan-¹(0.49)

(B) because truck moves constant speed in horizontal direction so,
acceleration in horizontal direction =0 but in vertical direction , acceleration= -9.8 m/s²
Hence, acceleration of stone = -9.8j m/s²

Answer 2

Answer:

A.20m/s

B.22.36m/s if value of g is 10.

if we take value of g is 9.8 then we obtain 22.4