A train starts from a station and moves with a constant acceleration for 2 minutes. If it covers a distance of 400m in this period, find the acceleration?
Answers
Answered by
154
hy friend here given that
initial velocity (u)= 0m/s
final velocity (v)=400
time(t)= 2 minutes
then,
Acceleration ( a)= v-u/t
= 400/2
= 200m/meter square
initial velocity (u)= 0m/s
final velocity (v)=400
time(t)= 2 minutes
then,
Acceleration ( a)= v-u/t
= 400/2
= 200m/meter square
sahul:
U should not give wrong answer
sorry by mistake i wrote another number
Can't u edit it
here isnt given thats why i couldnt edit
Ok I hv reported it
oh
Answered by
174
Answer:
Acceleration of the train = 0.0556 m/s^2
Explanation:
Given details:
Distance covered by the train (S)= 400 m.
Time taken for the travel (t) = 2 minutes = 2 * 60 seconds = 120 seconds.
Initial velocity = 0 m/s^2
Formula required:
We know that the equation of motion is
S = u*t + 1/2*a*t^2
Solution:
400 = 0*120 + 1/2 * a * 120^2
400 = 0 + a * 14400/2
400 = 7200*a
a = 400 / 7200
a = 0.0556
Acceleration = 0.0556 m/s^2
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