A man is moving with 36 kmph. The time of reaction is 0.9 seconds. On seeing an obstacle in the path, he applies brakes and decelerates at 5 m/s2, the total distance covered before he stops is:
The ratio of the heights from which two bodies are dropped is 3:5 respectively. The ratio of their final velocities is:
Velocity acquired by the two bodies falling from rest through a distance his given as v =
Given that :
A body starts to slide over a horizontal surface with an initial velocity of 0.2 m/s. Due to friction, its velocity decreases at the rate of 0.02 m/s2. How much time will it take for the body to stop?
If you whirl a stone on the end of the string and the string suddenly breaks, the stone will:
To describe the motion of an object we first specify a
The third equation of motion is
The third equation of motion is :-
( v² - u² = 2as)
u = initial velocity
v = final velocity
a = acceleration
s = distance
The time taken by a train to slow down from 80 kmh-1 to 20 kmh-1 with a uniform deceleration of 2 ms-2 is
v= 80*1000/3600 = 22.22 u=20*1000/3600=5.55 , Now a = v-u/t = 22.22-5.55/2 = 8.335
The equations of motion can be derived by using:
Consider a velocity - time graph for a uniformly accelerated body starting from rest is represented as follows.
u = velocity at time t1
v = velocity at time t2
If acceleration is represented as a, then, acceleration is defined as the rate of change in velocity.
⇒ a = v-u / t2 - t1
⇒ a = v-u /t
Or,v-u = at
⇒ v= u + at
A racing car has a uniform acceleration of 6 m/s2. In 10s it will cover:
Given:Acceleration = 6 m/s^2Time = 10 secAs car start from rest so intial velocity will be zero.u = 0S = ut + 1/2(at^2)S = 0 + 1/2 x 6 x 100S = 300 m.
A body performs an accelerated motion, with uniform speed. The motion of body is