Calculating Displacement and Distance for Vertical Motion Problems

[lionely]

Homework Statement

A ball is thrown vertically upwards and its height after t s is s m where
s= 25.2t - 4.9t²
Find the distance moved in the 3rd second

2. At the instant from which time is measured a particle is passing through O and traveling towards A along the straight line OA. It is s m from O after t s where
s= t(t-2)²

What is the particle's greatest displacement from O, and how far does it move, during the first 2 s?

The Attempt at a Solution

1. 25.2(3)-4.9(9) = 31.5m

Is the answer not 31.5m? But my book has a different answer.

2. For the greatest displacement I believe it is (32/27)m
but for the displacement after the first 2 seconds I'm not sure, cause if I put it in the formula for s I'll get 0.


Help is greatly appreciated.

 

[djh101]

1. What answer does your book have? The third second sounds like it could mean between t=2 and t=3, in which case the answer would be s₃ - s₂.

2. Greatest displacement should be the point where the particle stops and turns around, in other words, the point where the speed is 0, which should be 32/27. For the second part, I believe it is asking the overall distance traveled by the particle, not the net displacement (e.g. if you run around a 400m track, you will have run 400m but your net displacement will be 0 since you are at the same position you started at).

To find total distance travelled, find the critical points of your velocity (where s' = 0), plug t into s for each point, and take the sum of the differences between every two points. In other words:

Distance = (Distance b/w points 0 & 1) + (Distance b/w points 1 & 2) + ... + (Distance b/w points n-1 & n)
Distance = |S₁ - S₀| + |S₂ - S₁|

 

[lionely]

for 1 the answer is 2.5m

 

[Ray Vickson]

Homework Statement

A ball is thrown vertically upwards and its height after t s is s m where
s= 25.2t - 4.9t²
Find the distance moved in the 3rd second

2. At the instant from which time is measured a particle is passing through O and traveling towards A along the straight line OA. It is s m from O after t s where
s= t(t-2)²

What is the particle's greatest displacement from O, and how far does it move, during the first 2 s?

The Attempt at a Solution

1. 25.2(3)-4.9(9) = 31.5m

Is the answer not 31.5m? But my book has a different answer.

2. For the greatest displacement I believe it is (32/27)m
but for the displacement after the first 2 seconds I'm not sure, cause if I put it in the formula for s I'll get 0.


Help is greatly appreciated.

1) Your 31.5 m is the displacement at t = 3, not the distance moved between times t=2 and t=3.

2) Yes, s(2) = 0, but the particle travels out and back, so has traveled a finite, nonzero distance.

BTW: in 2), there is no greatest displacement from zero, since s(t) → ∞ as t → ∞. Perhaps the question put an upper bound on t?

 

[lionely]

No there's no upper bound that's all of the preamble to the question.

 

[djh101]

Okay. Yes, #1 wants the distance traveled from t=2 to t=3. The net displacement is .7m, but there is a critical point between 2 and 3 where s' = 0. As per my above post, first find the critical point and then add the absolute values of the distance traveled before and after the critical point.

 

[lionely]

I'm sorry I don't fully understand between the 2nd and 3rd second there's a point when displacement is 0? I should find that and then add to the .7?

 

[djh101]

No, there's a point where the velocity, s' (ds/dt), is zero. This is a critical point where velocity changes sign and the ball turns around (when the ball reaches its highest point and starts falling back towards earth). Plugging in t doesn't tell you how far the ball has traveled, it only tells you where the ball is at that point in time. To find out how far the ball actually traveled, you have to add up the distances that the ball traveled between the boundary and critical points.

|distance 1| + |distance 2| where distance 1 = s_{critical point} - s₂ and distance 2 = s₃ - s_{critical point}

 

[lionely]

So the I subtract the the values of s for when t=3 and t=2
I got from the max height(32.4m)

32.4-31.5(t=3) = 0.9 and 32.4-30.8(t=2)= 1.6
and then add them and that's the distance traveled 2.5m?

All of this makes the 3rd second? Hm..