A balloon starts rising from ground from rest with an upward acceleration of 2 m/s2. Just after 1 s ,a stone is dropped from it .What will be speed of stone soon after being dropped?
Answers
Answer:
Here, the driver of train A manages to avoid the collision by a constant retardation. Hence, the train A must be behind train B both moving in same direction.
Given that speed of A = 144 km h-¹ = 40 m s-¹
And that of train B = 108 km h-¹ = 30 m s-¹
Velocity of train A with respect to B =
= (40 - 30) ms-¹
= 10 ms-¹
The train eventually comes at rest to avoid the collision. So, v = 0 ms-¹
We have distance between them, s = 1 km = 1000 m (The train just travels this distance and stops to avoid the collision)
So we have
s = 1000 m
v = 0 ms-¹
u = 10 ms-¹
Putting them in third equation of motion in a straight line,
v² - u² = 2as
→ 0 - (10)² = 2a(1000)
→ -100 = 2000a
→ a = -1/20 ms-²
It is to be noted that this acceleration is relative acceleration between the two train. Now since train B is going on with constant speed and with no acceleration, this relative acceleration serves as the acceleration of train A.
We know that a = (v - u)/t
Hence,
-1/20 = (0 - 10)/t
→ -1/20 = -10/t
→ t = 200 seconds.
Hence, the retardation is -1/20 ms-² and is produced for 200 seconds.