Question

The acceleration function a(t) (in m/s2) and the initial velocity v(0) are given for a particle moving along a line.

a(t)=2t+4,

v(0)=-5

(a) Find the velocity v(t) at time t.

v(t)= ? m/s

(b) Find the total distance d traveled during the time interval given above.

d=? m

Answer 1

A:
`v(t)=-t^2+4t-5`

Explanation:

Acceleration is second derivative of position, velocity is first derivative. Therefore, the velocity is the integral of acceleration.

`v(t)=\int\ {2t+4} \, dt`

Integrate:

`-t^2+4t+C`

V(0)=-5:

`-0^2+4(0)+C=-5\\C=-5`

Therefore, v(t):

`v(t)=-t^2+4t-5`