Asked by Jessica
A ball rolls from rest down a smooth ramp that is 16.0m long. The acceleration of the ball is constant at 40ms^2.
A) how long does the ball take to roll the first 4.0m?
B) what is the speed of the ball when it is halfway down the ramp? (how far it has travelled)
C) how long does the ball take to travel the final 8.0m?
A) how long does the ball take to roll the first 4.0m?
B) what is the speed of the ball when it is halfway down the ramp? (how far it has travelled)
C) how long does the ball take to travel the final 8.0m?
Answers
Answered by
Damon
d = (1/2) a t^2 = 20 t^2
4 = 20 t^2
t^2 = .2
t = 0.447 seconds
4 = 20 t^2
t^2 = .2
t = 0.447 seconds
Answered by
Damon
8 = 20 t^2
t^2 = 0.4
t = .632
v = a t = 40 ( .632) = 16 m/s
t^2 = 0.4
t = .632
v = a t = 40 ( .632) = 16 m/s
Answered by
Jessica
which question is that for
A or C?
A or C?
Answered by
Jessica
What is the final speed of the ball?
Answered by
Damon
Vi at 8 m = 16
now accelerates at 40 m/s^2 for 8 more meters
v = Vi + a t = 16 + 40 t
x = Xi + Vi t +20 t^2
16 = 8 + 16 t + 20 t^2
20 t^2 + 16 t - 8 = 0
5 t^2 + 4 t - 2 = 0
t = .348 seconds
now accelerates at 40 m/s^2 for 8 more meters
v = Vi + a t = 16 + 40 t
x = Xi + Vi t +20 t^2
16 = 8 + 16 t + 20 t^2
20 t^2 + 16 t - 8 = 0
5 t^2 + 4 t - 2 = 0
t = .348 seconds
Answered by
Damon
Hey, it takes awhile :)
by the way
https://www.mathsisfun.com/quadratic-equation-solver.html
by the way
https://www.mathsisfun.com/quadratic-equation-solver.html
Answered by
Damon
They did not ask for final speed but I gave you the recipe:
v = Vi + a t = 16 + 40 t
= 16 + 40 ( .348)
v = Vi + a t = 16 + 40 t
= 16 + 40 ( .348)
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