Question

This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. The acceleration function (in m/s2) and the initial velocity v(0) are given for a particle moving along a line. a(t) = 2t + 4, v(0) = −32, 0 ≤ t ≤ 6

a) Find the velocity at time t.
b) Find the distance traveled during the given time interval

Answer 1

Answer:

(a) Velocity at time t will be  $v(t)=t^2+4t-32$

(B) Distance will be -48 m

Explanation:

We have given $a(t)=2t+4$

And $v(0)=-32$

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

So $v(t)=\int (2t+4)dt$

$v(t)=t^2+4t+c$

As $v(0)=-32$

So $-32=0^2+4\times 0+c$

c = -32

So $v(t)=t^2+4t-32$

(b) We have to find the distance traveled

So $s(t)=\int_{0}^{6}v(t)dt$

$s(t)=\int_{0}^{6}(t^2+4t-32)dt$

$s(t)=\int_{0}^{6}(\frac{t^3}{3}+2t^2-32t)$

$s=(\frac{6^3}{3}+2\times 6^2-32\times 6)-0=72+72-192=-48m$