# Homework Help: Motion related Questions

1. Apr 18, 2005

### Fiona

Hello All,

Would really appreciate your help, with the following questions :

• With positive value of acceleration, acceleration must be speeding up. If so why?

... I really didn't understand this question.
• A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed-time graph of its motion between t=0 to 12 seconds

For, this u= 0, s= -90, t = 12 seconds. So I use v2 = u2 + 2as? After doing that I get V. But how do I calculate the number of collisions?

• The speed time graph f a particle moving along a fixed direction is shown in the picture. Obtain the distance traversed by the article between ... (a) t = 0 s to 10 s and ... (b) 2 s to 6s

For the first instance its just 1/2 * b * h = 0.5 * 10 * 12 = 60 m
But for the second one, I don't know the value of speed (height). Therefore, how can I calculate area of the triangle for the distance?

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Last edited: Apr 18, 2005
2. Apr 18, 2005

### HallsofIvy

At first I didn't understand it either! I had no idea what "acceleration must be speeding up" meant- 'speed' and 'acceleration' are two different things. I would guess that it means "increasing". However the crucial part of this is "If so, why". This is really a "true or false" question. Suppose the acceleration is constant "9.81 m/sec2". That's positive but the the acceleration is not changing at all. Is the original statement true or false?

.
You need s= (1/2)(9.8)t2+ ut+ d and v= 9.8t+ u where u is the speed and d is the height at t=0 (since you took d= -90, your are taking positive distance to be down and s= 0 at the floor- that's why the acceleration is 9.8 rather than -9.8).
In this case, s= 4.9t2- 90. The first time the ball hits the floor, s= 0:
solve s= 4.9t2- 90= 0 to find the time the ball hits the floor. (That has two solutions. Obviously, you want the positive one.)
At that point, the speed is 9.8t+ u (using the t you just found) downward. After the ball bounces its speed will be .9 of that and upward. Set up the same s equation with initial speed u= the new speed, d= 0, and, again, set equal to 0 and solve for t to find the next time it bounces. (Again, that will have two solutions, one of them 0, you want the positive one. This is the time between bounces. The actual time from the initial drop to the second bounce is the sum of those two time.)
Okay, calculate the speed the second time it bounces (you should be able to see that it is the same as the speed upward after the first bounce, but downward), multiply by .9 to get the speed after the bounce, and repeat. Keep doing that until t reaches 12 seconds.

You are exactly right that the distance traversed by the article is the area of that triangle. To do
(b), note that 0 to 2 seconds is a small right triangle on the left and 6 to 10 seconds a right triangle on the right. Find the areas of those two and subtract from your answer to (a). How do you find the heights of those? The simplest way is to use "similar triangles". Looking just at the right triangle from t=0 to 5, we see that it base 5 and height 12. The little triangle from t= 0 to 2 has base 2 and height "x". Since the two triangles are similar (they have the same angles), x/2= 12/5. Solve for the height x and use that to find the area. The right half triangle (from t= 5 to 10) also has base 5 and height 12. The small triangle from t= 6 to 10 has base 4 and height "x". Since the two triangles are similar, x/4= 12/5. Again, solve for the height x and use that to find the area.

Last edited by a moderator: Apr 18, 2005
3. Apr 18, 2005

### Fiona

Hi,

Thanks a lot for your help! The similar triangles one was really easy once I got the concept right. The distance for the part (b) is 36m according to my calculations.

Regarding the 2nd question ...

How did you get s= 4.9t2- 90? Also you've used s= (1/2)(9.8)t2+ ut+ d .... but the equation is just s = 1/2 at2 + ut. If you introduced a variable d shouldn't it be added to the LHS too?

Once again. Thanks for your help!

Last edited: Apr 18, 2005
4. Apr 18, 2005

### HallsofIvy

. And where did you get that equation? s= (1/2)at2+ ut is correct IF s= 0 when t= 0. But you appear to be taking s= 0 to be at the floor and the intial height to be -90 m. If you use s= (1/2)a t2+ ut+ d, then, setting t= 0, the inital value of s is s= d. That's why I said, in this problem, with the initial height d= -90 and initial speed u= 0, you have s= (1/2)(9.8)t2- 90.

5. Apr 18, 2005

### Fiona

Hmm, Allright I see. I'll work on that a little later.

I was just doing this other question and got stuck with quite a few variables. I would be glad to know if you could help me eliminate a few.

Q: A stone is dropped from a height of 19.6 m. At the same time another stone is projected from vertically below the ground with sufficient velocity to enable it to ascend 19.6 m. Find when and where the two stones meet.

This is what I did:

Let one stone meet at h metres
Then, the other stones meets at (19.6 - h) metres.

... where h is the point where the two stones meet.

<<<<First Stone>>>

u=0, s = - (19.6 - h), a = -9.8

-19.6 + h= 0 + 1/2 * -9.8 * t2
-19.6 + h = 4.9t2 --(1)

<<Second Stone -complete drop>>

S = 19.6 m, v=0, a = -9.8

0 = u2 + 2 * -9.8 * 19.6
0 = u2 - (19.6)2
u = 19.6 m/s

<<second stone -- meeting at h>>

s = h, u = 19.6, a = -9.8

v2 = (19.6)2 + 2 * 9.8 * h ... (2)

... now I've got velocity and h in the same equation. Here's where I'm stuck.

6. Apr 18, 2005

### HallsofIvy

You haven't used the fact that the second stone is projected "with sufficient velocity to enable it to ascend 19.6 m". You can calculate the initial velocity necessary for that the be the maximum height the stone achieves. (By the way, "19.6" is exacty 2(9.8). That makes the arithmetic simpler.)

7. Apr 18, 2005

### Fiona

Hmm, I got it.

The time period is 1 sec and height at which they meet is 14.7 m.

Thanks for the help.

8. Apr 18, 2005

### HallsofIvy

Yes, that's very good. Notice that that is exactly 3/4 of the original 19.6 m height!

9. Apr 18, 2005

### Fiona

Hmm, yup I hadn't thought of that before :).