Question

A man start moving with initial velocity 12ms and the uniform acceleration is 2ms. If the distance travelled by the man in the last second of its motion is 27m, then the time taken by the man in his journey is​

Answer 1

`8\; {\rm s}`.

Step-by-step explanation:

To solve this question, start by finding the final velocity from the distance travelled in the last second of the motion. After that, divide the velocity gained during the entire motion by acceleration to find the duration of the motion.

Let
`v\; {\rm m\cdot s^(-1)}` be the final velocity.

The velocity at the beginning of the last second (
`\Delta t = 1\; {\rm s}`) would be
`(v - a\, \Delta t) \; {\rm m\cdot s^(-1)}`, which is equal to
`(v - 2)\; {\rm m\cdot s^(-1)}`. Since acceleration is constant, average velocity in that period of
`\Delta t = 1\; {\rm s}` would be:

`\begin{aligned} (\text{average velocity}) &= \frac{(\text{velocity at start}) + (\text{velocity at end})}{2} \\ &= \frac{(v - 2)\; {\rm m\cdot s^(-1)} + v\; {\rm m\cdot s^(-1)}}{2} \\ &= (v - 1)\; {\rm m\cdot s^(-1)}\end{aligned}`.

Multiply the average velocity by the duration to find an expression for the distance travelled within that
`\Delta t = 1\; {\rm s}`:

`\begin{aligned}(\text{distance}) &= (\text{average velocity}) \, (\text{time taken}) \\ &= ((v - 1)\; {\rm m\cdot s^(-1)})\, (1\; {\rm s}) \\ &= (v - 1)\; {\rm m}\end{aligned}`.

Given that the distance travelled in that period of time is
`27\; {\rm m}`:

`(v - 1) = 27`.

`v = 28`.

In other words, the final velocity of the person is
`28\; {\rm m\cdot s^(-1)}`.

It is given that the initial velocity is
`u = 12\; {\rm m\cdot s^(-1)}` and acceleration is
`a = 2\; {\rm m\cdot s^(-2)}`.

In other words, velocity has increased from
`u = 12\; {\rm m\cdot s^(-1)}\!` to
`v = 28\; {\rm m\cdot s^(-1)}` at a rate of
`a = 2\; {\rm m\cdot s^(-2)}`. To find the time required, divide the change in velocity by the rate of change:

`\begin{aligned}t &= (v - u)/(a) \\ &= \frac{(28\; {\rm m\cdot s^(-1)}) - (12\; {\rm m\cdot s^(-1)})}{(2\; {\rm m\cdot s^(-2)})} \\ &= 8\; {\rm s}\end{aligned}`.

In other words, the time required was
`8\; {\rm s}`.