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Vectors and Acceleration in a Straight Line

  • Thread starter Peter G.
  • Start date
  • #1
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What is the change in velocity when:
a) 6.0 m/s becomes - 15 m/s
b) 5.0 m/s East becomes 15 m/s West

My doubt is mainly the notation: How I should answer, here is my try:

a) -9 m/s with 180 degrees change in direction
b) 9 m/s with from direction bearing 90 degrees to 270 degrees

And: A baby buggy rolls down a ramp which is 15 m long. It starts from rest and accelerates uniformly and takes 5.0 seconds to reach the bottom:
a) Average Velocity: 15 /5 = 3 m/s

Now, b asks for the velocity at the bottom. I don't know how to calculate the velocity without the acceleration, which is asked in question c)

c) s = ut x 1/2at^2
a = 1.2 m/s^2

and then

b) v = u + at
v = 6 m/s

Anyone can teach me how to get the velocity without the acceleration and with the notation in the first question?

Thanks,
Peter
 

Answers and Replies

  • #2
verty
Homework Helper
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198
You can find the final velocity with this formula (only when acceleration is constant):

s = t(u+v)/2
 
  • #3
442
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Aw, yea that's right, thanks.

And what about my answers to the first question? Think they are O.K?
 

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