Question

A train is moving with an initial velocity of 30m/sec . The brakes are applied So as to produce a uniform acceleration of -1.5m/sec calculate the time in which It will come to rest

Answer 1

Answer:

20 seconds

Explanation:

Given:

• Initial velocity of the train = u = 30 m/s

• Acceleration produced in the train = -1.5 m/s²
• Final velocity = v = 0 m/s

To find:

• Time taken for the train to come to rest

Using first equation of motion which says:

V=u+at

0=30-1.5×t

-30=-1.5t

t = $\dfrac{-30}{-1.5}$

t = $\dfrac{30}{1.5}$

t = 20 seconds

The time taken for the train to come to rest is equal to 20 seconds

Answer 2

Given :

• A train is moving with velocity of 30 m/s
• It comes to rest after applying breaks
• Acceleration is - 1.5 m/s²

To Find :

• Time in which it will come to rest

Explanation :

We are given that the train is moving with the velocity of 30 m/s, it means initial velocity (u) is here 30 m/s. And after travelling some distance breaks are applied it means that the train comes to rest, So here final velocity (v) will be 0 m/s. The retardation or negative acceleration which means acceleration is applied in opposite direction of motion is given as - 1.5 m/s².

Here, we will use simply Kinematics equations or say Equations of motion.

Solution :

Use 1st equation of Kinematics or Motion :

$\implies \sf{v \: = \: u \: + \: at} \\ \\ \implies \sf{0 \: = \: 30 \: + \: (-1.5 \: \times \: t) } \\ \\ \implies \sf{-30 \: = \: -1.5t} \\ \\ \implies \sf{1.5t\: = \: 30} \\ \\ \implies \sf{t \: = \: \dfrac{30}{1.5}} \\ \\ \implies \sf{t \: = \: 20 }$

$\therefore$ Time required by train to come to rest is 20 seconds