Given, the initial velocity (u) of the train = 0m.s-1 (at rest)
Terminal velocity (v) of the train = 40km/hour = 11.11 m.s-1
Time interval, t = 10 minutes = 600 s.
The acceleration of the train is given by a \(= \frac{v-u}{t}\) \( =\frac{11.11\,m.s^{-1}-0}{600s}\) \( =0.0185\,m.s^{2}\)
Answered by Vishal kumar | 1 year agoA bus starting from rest moves with a uniform acceleration of 0.1 m s-2 for 2 minutes. Find
(a) the speed acquired,
(b) the distance traveled.
A farmer moves along the boundary of a square field of side 10m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?
What is the quantity which is measured by the area occupied below the velocity-time graph?
A bus starting from rest moves with a uniform acceleration of 0.1 m s-2 for 2 minutes. Find
(a) the speed acquired,
(b) the distance travelled.
What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?