Describe the motion of a body which is accelerating at a constant rate of 10ms-2. If the body starts from rest ; how much distance will it cover in 2 seconds
Answers
Answered by
1
It is uniformly accelerated motion.
S = ut + 0.5at^2
= 0 + 0.5*10*2^2
= 20 m
Distance travelled by body in first 2s is 20 m
S = ut + 0.5at^2
= 0 + 0.5*10*2^2
= 20 m
Distance travelled by body in first 2s is 20 m
Answered by
0
A body starts from rest means initial velocity is 0 m/s and travelling with an acceleration of 4 m/s².
We have to find the distance travelled by the body in the 4th second.
We know that distance covered by body in nth second is given by-
Sn = u + a/2 (2n - 1)
Here; n = 4 sec, u = 0 m/s and a is 4 m/s².
Let us assume that the distance covered in the 4th second is x km.
Substitute the known values in the above formula,
→ x = 0 + 4/2 (2*4 - 1)
→ x = 0 + 2 (8 - 1)
→ x = 0 + 2(7)
→ x = 0 + 14
→ x = 14
As, the x is the distance covered in 4th second.
Therefore, the distance covered by the body in the 4th second is 14 m.
Similar questions