A ball with an initial velocity of 10 m/s is rolling up a hill. The ball accelerates at a rate of 2.9 m/s/s.
How much time will it take for the ball to be 12 m from the starting position? Note that there are two possible answers here.
The first time the ball is 12 m from the starting position will be _____
The second time the ball is 12 m from the starting position will be _____
work would be nice :)
How much time will it take for the ball to be 12 m from the starting position? Note that there are two possible answers here.
The first time the ball is 12 m from the starting position will be _____
The second time the ball is 12 m from the starting position will be _____
work would be nice :)
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As things slow down when rolling up a hill, I guess you mean acceleration = -2.9 m/s/s (taking the direction of initial velocity as positive, acceleration will be negative).
s = ut + ½at² (or x = vi.t + ½at² if you use those symbols)
12 = 10t + ½ x (-2.9) x t²
1.45t² -10t + 12 = 0
Solve the quadratic in the normal way
t = [10 ± √(10² - 4x1.45x12) ] / (2x1.45)
= (10 ± 5.51) / 2.9
= 1.5s and 5.3s
s = ut + ½at² (or x = vi.t + ½at² if you use those symbols)
12 = 10t + ½ x (-2.9) x t²
1.45t² -10t + 12 = 0
Solve the quadratic in the normal way
t = [10 ± √(10² - 4x1.45x12) ] / (2x1.45)
= (10 ± 5.51) / 2.9
= 1.5s and 5.3s
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V = 10 m/s
a = 2.9 m/s
V =at solve for t
t=10/2.9 = 3.44 seconds. So the ball will stop in 3.44 seconds. The average velocity is 5 m/s
If D= 12 then t=d/v(avg) = 12/5 = 2.4 seconds. This is the time it takes to get 12 m from start.
Lets figure out how far the ball rolls before it stops. We know the trip takes 3.44 seconds with an average velocity of 5 m/s so the total distance is 17.2 meters. Taking 3.44 seconds.
Now lets go from where it stops to the 12M point. 17.2 - 12 = 5.2 The ball has to travel back 5.2 meters from it's stopped point at the top of the hill. But how long does this take? Use the same acceleration of 2.9 m/s/s
Call V the average velocity and v is the instantaneous velocity
D = 5.2
a= 2.9
V=d/t
v=at so V= (1/2 v) so (1/2)v = D/t now substitute it into the V=at equation:
or 1/2 at = D/t multiply both sides by t: (1/2)att=D 1/2tt=D/a.
Solve for t so t(squared) = 2D/a or 2(5.2)/2.9= 3.586
now take the square root of 3.586 therefor t=1.89 seconds to roll back down to the 12M point. Add the 3.44 for a total time of 5.29 seconds.
a = 2.9 m/s
V =at solve for t
t=10/2.9 = 3.44 seconds. So the ball will stop in 3.44 seconds. The average velocity is 5 m/s
If D= 12 then t=d/v(avg) = 12/5 = 2.4 seconds. This is the time it takes to get 12 m from start.
Lets figure out how far the ball rolls before it stops. We know the trip takes 3.44 seconds with an average velocity of 5 m/s so the total distance is 17.2 meters. Taking 3.44 seconds.
Now lets go from where it stops to the 12M point. 17.2 - 12 = 5.2 The ball has to travel back 5.2 meters from it's stopped point at the top of the hill. But how long does this take? Use the same acceleration of 2.9 m/s/s
Call V the average velocity and v is the instantaneous velocity
D = 5.2
a= 2.9
V=d/t
v=at so V= (1/2 v) so (1/2)v = D/t now substitute it into the V=at equation:
or 1/2 at = D/t multiply both sides by t: (1/2)att=D 1/2tt=D/a.
Solve for t so t(squared) = 2D/a or 2(5.2)/2.9= 3.586
now take the square root of 3.586 therefor t=1.89 seconds to roll back down to the 12M point. Add the 3.44 for a total time of 5.29 seconds.
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we can use
dist = v0 t + 1/2 a t^2
where dist = 12m
v0=10m/s
a=-2.9m/s/s and we solve the quadratic
12 = 10t -1.45t^2
or 1.45t^2 - 10t+12=0
t=1.55 and 5.35s
dist = v0 t + 1/2 a t^2
where dist = 12m
v0=10m/s
a=-2.9m/s/s and we solve the quadratic
12 = 10t -1.45t^2
or 1.45t^2 - 10t+12=0
t=1.55 and 5.35s