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

Answer:

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

Step-by-step 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):

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