Physics Motion Problem (Displacement)

In summary, the question is whether the boy will catch the ball before the end of the 60-meter ramp, given that the ball starts accelerating at 0.2 m/s(squared) and the boy starts accelerating 12 seconds later at 0.9 m/s(squared). The attempt to solve this using the equations v2=v1 + aΔt and Δd=v1Δt+0.5a(Δt)squared was unsuccessful, and graphing was not allowed. The solution involves finding the time when the displacement of the boy and the ball equals 60 meters, and it is possible for the boy to catch the ball if this time is less than or equal to 60 seconds.
  • #1
lp03269
2
0
1. Hi, my question deals with displacement and given acceleration : A boy and a ball are at the top of a ramp. the ball rolls down accelerating at 0.2 m/s(squared). The boy chases after the ball 12 seconds later accelerating at 0.9 m/s(squared). The length of the ramp is 60 meters. Will the boy catch the catch the ball before the end of the ramp, if so what displacement/ time will it be at


2. I was trying to use v2=v1 + aΔt, and manipulate it for the getting the time by after using the equation, Δd=v1Δt+0.5a(Δt)squared



3. My attempt was by trying to divide the displacement by the acceleration and i think i got time(squared) so i square rooted that and than i plugged in the that time by the acceleration to get velocity at that time, and than calculating the displacement by multiplying the velocity by the time. I also tried to substitute the time/ displacement values into each other but i could not balance it out to solve it algebraically. I also tried graphing it and i got my answer somewhere around 15 seconds but I'm not allowed to graph

Any Help would be appreciated... Thanks
 
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  • #2
the total length of ramp = 60 m
So, s= 1/2(0.2)t^2 =0.1t^2
after 12s s' = (0.1)(12)^2=14.4 m
14.4+(0.1)t^2
for the motion of boy
s=1/2(0.9)t^2
combine the equation, find t
if 14.4+(0.1)t^2>=60 it is impossible
<=60 , it is possible
 
  • #3


Hi there,

Thank you for your question. In order to solve this problem, we can use the equations of motion in one dimension. The first equation we can use is d = v0t + 1/2at^2, where d is the displacement, v0 is the initial velocity, a is the acceleration, and t is the time.

We know that the ball has a constant acceleration of 0.2 m/s^2, so we can use this equation to find the displacement of the ball after 12 seconds:
d1 = (0 m/s)(12 s) + 1/2 (0.2 m/s^2)(12 s)^2 = 144 m

Next, we can use the same equation to find the displacement of the boy after 12 seconds, but with an initial velocity of 0 m/s and an acceleration of 0.9 m/s^2:
d2 = (0 m/s)(12 s) + 1/2 (0.9 m/s^2)(12 s)^2 = 64.8 m

Since the ramp is 60 meters long, we can see that the boy has not caught up to the ball yet. We can also use the equation v = v0 + at to find the final velocity of the boy after 12 seconds:
v2 = (0 m/s) + (0.9 m/s^2)(12 s) = 10.8 m/s

Now, we can use the equation d = v0t + 1/2at^2 again to find the time it takes for the boy to catch the ball. We know the displacement is 60 m and the initial velocity is 10.8 m/s. So, we can rearrange the equation to solve for t:
60 m = (10.8 m/s)t + 1/2 (0.9 m/s^2)t^2
0.45t^2 + 10.8t - 60 = 0

Using the quadratic formula, we can solve for t and get two solutions: t = 5.54 s or t = -21.54 s. Since time cannot be negative, we can discard the negative solution and conclude that it will take the boy 5.54 seconds to catch the ball.

Therefore, the boy will catch the ball at a displacement of 60 meters after 5.54 seconds. I hope
 

Related to Physics Motion Problem (Displacement)

1. What is displacement in physics?

Displacement is a measurement of the distance and direction an object has moved from its initial position to its final position. It is a vector quantity, meaning it has both magnitude (size or amount) and direction.

2. How is displacement different from distance?

Distance is the total length of the path an object has traveled, while displacement is the shortest distance between an object's initial and final positions, taking into account the direction of movement. Distance is a scalar quantity, meaning it only has magnitude, while displacement is a vector quantity.

3. What is the formula for calculating displacement?

The formula for displacement is Δx = xf - xi, where Δx is the displacement, xf is the final position, and xi is the initial position. This formula can also be written as Δx = xf - xi.

4. Can displacement be negative?

Yes, displacement can be negative. This indicates that the object has moved in the opposite direction from its initial position. A positive displacement means the object has moved in the same direction as its initial position.

5. What is the unit of measurement for displacement?

The unit of measurement for displacement is the same as that for distance, which is typically meters (m) in the metric system and feet (ft) in the imperial system. However, displacement is a vector quantity, so it should be written with a direction, such as "5 meters east" or "10 feet south."

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