Question

A particle is moving along a straight line and the acceleration is given by a(t)=3t+5, (units m/s2). a)Find v(t) , the velocity at time t if the initial velocity is v(0) = -4 ft/s? b)Find the total distance traveled during the time interval [0, 2]?

Answer 1

Velocity will be
`v(t)=(3t^2)/(2)+5t-4ft/sec`

Distance traveled = 6 feet

Step-by-step explanation:

We have given acceleration
`a(t)=3t+5`

We know that
`v(t)=\int a(t)dt`

So
`v(t)=\int 3t+5dt`

`v(t)=(3t^2)/(2)+5t+c`

We have given
`v(0)=-4ft/sec`

So
`-4=(3* 0^2)/(2)+5* 0+c`

c = -4 ft/sec

So
`v(t)=(3t^2)/(2)+5t-4ft/sec`

Now we have to find distance traveled in interval [0,2]

So distance
`s=\int v(t)dt`

`s=\int_(0)^(2)v(t)dt`

`s=\int_(0)^(2)((3t^2)/(2)+5t-4)dt`

`s=((t^3)/(2)+(5t^2)/(2)-4t)_(0)^(2)`

`=(4+10-8)-0`

`=6ft`