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Sports Car Acceleration Question
Answers and Replies
You need to use the kinematic equations v=u+at v2=u2+2as s=ut+1/2at2 Which one do you think you will have to use?
What does the u stand for, that's not in my book?
The car initially had a velocity of 110m/5s (22m/s) and came to a stop (0m/s) in 4.0s. The acceleration is the change in velocity over a time interval.
So would we use the v=u+at since I have the 4.0s and the 22m/s? but what would go in for the u?
u is initial velocity and v is final velocity.
Okay I figured it out thanks so much the answer was -5.5 m/s^2 and then in g's it was -0.56 g's
take u (initial speed) = 110 / 5 = 22m/s v = u + at Suggested for: Sports Car Acceleration Question
Pau L. a. How fast is the far moving during the first 5 s? b. What is the magnitude of the acceleration of the car as it brakes and comes to a stop? More 1 Expert Answer
a. You are given distance and time, and velocity is the ratio of the two. b. The car goes from a velocity of V1 to velocity of V2 in a given time. Acceleration is the rate of change of velocity, so, a = (V1 - V2) / (t1 - t2) Note: This is a very simple question that for AP Physics should have been well demonstrated in the textbook. Reread the textbook and your class notes to see where this is explained. Still looking for help? Get the right answer, fast.ORFind an Online Tutor Now Choose an expert and meet online. No packages or subscriptions, pay only for the time you need. |
A sports car moving at a constant velocity travels 120
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