Question

A train starts from rest and accelerates uniformly to reach a velocity of 30m/s in 4s. It then maintains a constant speed for 4s and then uniformly retarded and comes to rest in 2s.Calculate (i)maximum velocity reached,(ii) acceleration in first 4s and last 2s.(iii)Total distance travelled (iv)Average velocity of the train."​

Answer 1

Given above is the graphical visualisation of the motion of the train. So that it becomes easy to understand.

So, in the first question it is asking the maximum velocity. Obviously, its 30m/s as you can see from the graph.

2nd question asks for the acceleration in the first 4 secs and last 2 secs.

In the first 4 seconds, we can calculate the acceleration using the formula

$v - u \div t$

which is 30/4 which results in 7.5m/s^2.

And in the same way we calculate for the last 2 seconds also. We get - - 15m/s^2 (deceleration).

Third question is the total distance travelled. We can calculate it directly from the graph in an easy way.

You can see that the graph looks like a trapezium. So the distance can be calculated by finding the area of the trapezium which is

$h \times (a + b) \div 2$

Which is 30 × (4+10) /2 = 210 metres.

The last question is for finding the average velocity.

It can be calculated by the following steps. - :

v+u/2 in the 1st triangle + v+u/2 in the rectangle + v+u/2 in the third triangle.

Which is 15 + 30 + 15 = 60 m/s.

Where, The formula used for average velocity is v + u /2.

Hope it helps :)))