Vectors and Acceleration in a Straight Line

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SUMMARY

The discussion focuses on calculating changes in velocity and acceleration in physics, specifically addressing two scenarios: a change from 6.0 m/s to -15 m/s and from 5.0 m/s East to 15 m/s West. The correct answers are -9 m/s with a 180-degree direction change for the first scenario and 9 m/s with a direction change from 90 degrees to 270 degrees for the second. Additionally, the average velocity of a baby buggy rolling down a 15 m ramp in 5 seconds is calculated as 3 m/s, while the final velocity is determined using the formula v = u + at, resulting in 6 m/s with an acceleration of 1.2 m/s².

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  • Understanding of basic kinematics concepts, including velocity and acceleration
  • Familiarity with the equations of motion, specifically v = u + at
  • Knowledge of vector notation and direction in physics
  • Ability to calculate average velocity and final velocity in uniformly accelerated motion
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  • Study the equations of motion in detail, focusing on constant acceleration scenarios
  • Learn about vector addition and subtraction in physics
  • Explore practical applications of kinematics in real-world scenarios
  • Review problems involving acceleration and velocity to reinforce understanding
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Students studying physics, educators teaching kinematics, and anyone interested in understanding motion and acceleration principles.

Peter G.
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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
 
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You can find the final velocity with this formula (only when acceleration is constant):

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

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

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