Question

A train started from rest and moved with constant acceleration. At one time it was traveling 21 m/s, and 180 m farther on it was traveling 45 m/s. Calculate (a) the acceleration, (b) the time required to travel the 180 m mentioned, (c) the time required to attain the speed of 21 m/s, and (d) the distance moved from rest to the time the train had a speed of 21 m/s.

Answer 1

Answer:

Part a)

acceleration is 4.4 m/s/s

Part b)

time taken by it 5.45 s

Part c)

Time taken to reach the initial speed is 4.77 s

Part d)

distance moved by the train is 50.11 m

Explanation:

Part a)

As we know that train is moving uniformly

so we will have

$v_f^2 - v_i^2 = 2 a d$

Now we have

$45^2 - 21^2 = 2a(180)$

$a = 4.4 m/s^2$

Part b)

In order to find the time we have

$v_f = v_i + at$

$45 = 21 + 4.4 t$

$t = 5.45 s$

Part c)

Time to reach the speed of 21 m/s is given by

$v = v_i + at$

$21 = 0 + 4.4 t$

$t = 4.77 s$

Part d)

Distance moved by the train is given as

$d = v_i t + \frac{1}{2}at^2$

$d = 0 + \frac{1}{2}(4.4)(4.77^2)$

$d = 50.11 m$

#Learn

Topic : Kinematics

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