A particle is moving in a straight line under constant acceleration alpha. If the intial velocity is `v_(0)`, sketch the v-t and s-t graphs. `(v_(0)gt0)`.

Explanation

v-t graph:
`v=u+at=v_(0)+at` ,
`t=0, v=v_(0)`
Shape: `y=v_(0)+ax`(equation of straight line)
`y=c+mx`
`c=v_(0)`, i.e. +ve, the line will cut the y-axis above the origin.
`m=alpha`, i.e. +ve, slope is +ve, so the inclination of line, `thetalt90^(@)`.

Since time cannot be -ve, the line will not be to the left of the origin.

s-t graph:
`s=ut+1/2 at^(2)=v_(0) t+(alpha t^(2))/2`
`t=0, s=0`
shape: `y=v_(0) x+(alphax^(2))/2` (parabola)
(a) The coefficient of `x^(2)` is `alpha/2`, i.e. +ve, so the parabola will be open upward.
(b) `y=v_(0)x+(alphax^(2))/2=x(v_(0)+(alphax)/2)=0`
`x=0, x=-(2v_(0))/alpha`
The parabola will cut the x-axis at `x=0, -(2v_(0))/alpha`.

But time cannot be -ve, hence, the s-t graph will be