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Old Sep 21st 2008, 05:37 PM   #1
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Velocity/Displacement Questions

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?

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34. A record of travel along a straight path is as follows:

1. Start from rest with a constant acceleration of 2.77 m/s2 for 15.0s
2. Maintain a constant velocity for the next 2.05 min.
3. Apply a constant negative acceleration of -9.47 m/s2 for 4.39 sec.

(a) What was the total displacement for the trip?
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Old Sep 24th 2008, 01:52 PM   #2
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do you not have a kinematics table in your text if so list your knowns and use the approiate equation
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Old Sep 24th 2008, 05:08 PM   #3
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Originally Posted by Blodwynne View Post
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
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