Question

A particle starts from s=0 and travels along a straight line with v=(t^2 - 4t + 3) m/s, where t is in seconds. Construct the v-t and a-t graphs for the time interval 0<=t<=4s.

Answer 1

Given that

`v = (t^(2) - 4t +3)`

the time interval is

`0\leq t\leq 4s`

We calculate the time

`v =t^(2) -4t +3`

`a = (dv)/(dt) = (d)/(dt)(t^(2) -4t +3)`

`a = 2t - 4`

`t = 2s`

Now, the graph of v-t for time interval
`0\leq t\leq 4s`

`v =t^(2) -4t +3`

at
`t = 0\ s , v = 0 - 0 + 3 = 3 m/s`

at
`t = 2\ s , v = 4 - 8 +3 = -1 m/s`

at
`t = 4\ s , v = 16 - 16 +3 = 3 m/s`

It is show in figure 1

Now, the graph of a-t for time interval

`v =t^(2) -4t +3`
`a = (dv)/(dt) = (d)/(dt)(t^(2) -4t +3)`

`a = 2t - 4`

at
`t = 0\ s , a = - 4 m/s^(2)`

at
`t = 2\ s , a = 0 m/s^(2)`

at
`t = 4\ s , a = 4 m/s^(2)`

It is show in figure 2

Hence, this is the required solution