Question

The acceleration function (in m/s^2) and the initial velocity are given for a particle moving along a line. Find (a) the velocity at time t and (b) the distance traveled during the given time interval.
a(t)= t+4, v(0)=5, 0 is less than or equal to t which is less than or equal to 10

I know the velocity is (t^2)/2 +4t +5 But whenever I start to integrate I always get 583 1/3. The answer sheet says that the answer is 416 2/3. Can anyone tell me what I'm doing wrong

Answer 1

There's a difference between "distance" and "displacement" (also known as position). In the usual context, the total "distance" traveled is the *absolute* distance, i.e. \(d(t)=\displaystyle \int_a^b |v(t)|\,dt\), whereas "displacement" is the *net* distance, i.e. \(s(t)=\displaystyle \int_a^b v(t)\,dt\).

Answer 2

I thought that too at first. But when you plug the velocity into the calculator both zeros are negative. So only the positive part of the absolute value works.

Answer 3

Alright then, it must be some algebraic error in your work. I'm getting the correct the answer in Mathematica.

Answer 4

\[d=\int_0^{10}\left(\frac{1}{2}t^2+4t+5\right)\,dt=\left[\frac{1}{6}t^3+2t^2+5t\right]_0^{10}=\frac{1000}{6}+200+50=\frac{1250}{3}\]

Answer 5

Oh I see what I did wrong I forget to multiply the antiderivative of t^2 by 1/2 Thank you so much!

Answer 6

You're welcome!