Originally Posted by **Blodwynne** 30. A truck on a straight road starts from rest and accelerates at 2.0 m/s2 until it reaches a speed of 20 m/s. Then the truck travels for 20s at a constant speed until the brakes are applied, stopping the truck in a uniform manner in an additional 5.0 s.
(a). How long is the truck in motion? **35 seconds**
(b) What is the average velocity of the truck during the motion described? |

a) The truck accelerates from rest at an acceleration of 2.0 m/s^2 until it attains a speed of 20 m/s. We need to find out how long this way. So

$\displaystyle v = v_0 + at$

$\displaystyle 20 = 2.0t$

$\displaystyle t = 10~s$

It travels with a constant speed of 20 m/s for 20 s.

Then it breaks to a stop for 5 s.

Thus the total trip took 10 + 20 + 5 = 35 s.

b) The truck accelerates from rest at an acceleration of 2.0 m/s^2 until it attains a speed of 20 m/s. We want to know how far the truck traveled in this time. So...

$\displaystyle v^2 = v_0^2 + 2ax$

$\displaystyle 20^2 = 2(2.0)x$

$\displaystyle x = \frac{20^2}{4} = 100~m$

Then the truck travels at a constant speed of 20 m/s for 20 s. This covers a distance of

$\displaystyle x = vt = 20 \cdot 20 = 400~m$

Then the truck decelerates to a stop in 5 s. Thus

$\displaystyle x = \frac{1}{2}(v_0 + v)t$

$\displaystyle x = \frac{1}{2}(20 + 0) \cdot 5 = 50~m$

So the total trip was 100 + 400 + 50 = 550 m long.

$\displaystyle \Delta v = \frac{550~m}{35~s} = 15.714~m/s$

-Dan