A partical is moving in a straight line under constant acceleration `4 m//s^(2)`. The motion starts from rest. Sketch v-t and s-t graph for the duration `t=0` to `t=5 s`.

Explanation

`(tou=0 " " toa=4m//s^2)/O`
To sketch graphs, we require some values: initial value, final value and values, at some critical points where change is taking place.
At `t=0, velocity=0`
`v=u+at=0+4t=4t`
At `t=5 s, v=20 m//s`

Now decided the shape of the graph.
`v=4t`
Write the quantity on the y-axis as `y`.
Write the quantity on the x-axis, as `x`.
`y=4x`(Equation of straight line)
Comparing it with `y=mx+c`, we get `m=4, c=0`.
`m=4`, i.e. +ve, so `thetalt90^(@), c=0`, line will pass through the origin.

s-t graph:
`s=ut+1/2 at^(2)=0+1/2xx4xxt^(2)=2t^(2)`
`t=0, s=0`
`t=4 s, s=32 m`
`s=2t^(2)`
`y=2x^(2)`(parabola, open upward)