The velocity-time graph is given below

Draw the acceleration-time graph and position-time graph, given that at t = 0, particle is at origin.

Explanation

Given that `a=(dv)/(dt)=tan 135^(@)=-1`
then,
again, `(dv)/(dt)=-1 rArr dv=-1dt`
`int_(4)^(v)dv=-int_(0)^(t)dt`
`(v)_(4)^(v)=-(t)_(0)^(t)`
`v-4=-t`
`v=4-t`
`v=(dx)/(dt)=4-t`
`dx=(4-t)dt=4dt-tdt`
`int_(0)^(x)dx=int_(0)^(t)4dt-int_(0)^(t)tdt`
`(x)_(0)^(x)=4(t)_(0)^(t)-((t^(2))/(2))_(0)^(t)`
`x-0=4(t-0)-(t^(2))/(2)=4t-(t^(2))/(2)`
`x=4t-(t^(2))/(2)`