Question

A cameraman on a pickup truck is traveling westward at 23 km/h while he videotapes a cheetah that is moving westward 30 km/h faster than the truck. Suddenly, the cheetah stops, turns, and then run at 46 km/h eastward, as measured by a suddenly nervous crew member who stands alongside the cheetah's path. The change in the animal's velocity takes 2.0 s. What are the (a) magnitude of the cheetah's acceleration according to the cameraman and the (b) magnitude of the cheetah's acceleration according to the nervous crew member?

Answer 1

The math doesn't seem like it would be all that hard, but I guess I am having problems conceptually for some reason, I feel rather silly.

Answer 2

The magnitude of the acceleration is just the magnitude of the velocity change divided by the time over which it happens. What are the initial and final velocities according to the guy on the truck?

Answer 3

The initial velocity should be 30km/h - 23km/h= 7km/h
Final velocity should be 30km/h + 23km/h= 53km/h

Right?

Answer 4

No. The initial velocity of 30 km/h is measured relative to the truck -- so the initial velocity is 30 km/h west.
The final velocity of 46 km/h east is measured relative to the ground, so the final velocity is 46+23 = 69 km/h east.

The magnitude of the velocity change is 30 + 69 = 99 km/h

Answer 5

For the guy on the ground, the initial velocity would be 30 + 23 = 53 km/h west. The final velocity would be 46 km/h east, meaning the magnitude of the velocity change would be 53+46 = 99 km/h.

Answer 6

Oh, sorry... I miss read the problem. I meant the 46 not 30, silly error.
It makes sense that it is going 30km/h faster relative to the truck.

Answer 7

So the magnitude of the velocity change is the same for both people?

Answer 8

Yes, meaning they would see the same acceleration as well. As they certainly must, right?

Answer 9

If the change of velocity is the same for both of them, and it takes the same amount of time, then it should I would think.

Answer 10

So... would the magnitude of the change in the cheetahs acceleration be 13.75 m/s^2?

Answer 11

Yes

Answer 12

And this is for part (a), in relation to the man on the truck? So how would the problem change for the crew member on the ground? It must somehow, or they wouldn't break the problem up... right?

Answer 13

Don't assume that the structure of the problem has anything to do with the answer. The procedure would not change at all.

Answer 14

Ok, so... If the magnitude of change in velocity is the same for both crew members, and the time that it takes is the same, does that mean that both answers would be the same?

Answer 15

Yes

Answer 16

Thank you very much, I think that I was reading into it too much. I really appreciate the help!

Answer 17

Sure. This is a specific example of an important principle -- if the observer is moving at a constant velocity, the acceleration cannot depend on that velocity at all.

Answer 18

Because it isn't the object that is having a change in velocity right?

Answer 19

I know its been solved but here is a workout of the math.

\[v_{ab} = v_{a}-v_{b}\]
That is velocity of a relative to b

In the case of the nervous crew member, they observe the cheetah travelling 53km/h west (truck speed + cheetah speed difference). After stopping and heading east at 46 km/h the crew member has observed a chance in speed of:

\[\Delta v1=(23 + 30) km/h (W) - 0 = 53 km/h (W)\]
\[\Delta v2 = 46 km/h (E) - 0 = 46 km/h (W)\]
These correspond to the velocity of the cheetah relative to the ground crew member.
Zero is the velocity of the ground crew member
\[\Delta v =53 km/h (W) - 46 km/h (E) = 99 km/h (W)\]
over a period of two seconds. Quickly converting to meters per second:
\[99\frac{ km }{h } \div 3.6 (\frac{ km*s }{ m*h }) = 27.5 m/s\].
\[a = \frac{ \Delta v }{ \Delta t } = \frac{ 27.5 m/s }{ 2.0s } = 13.75 \frac{ m }{ s^{2} }\]

Now for the cameraman.

The cameraman observes the cheetah having a relative velocity of 30 km/h (W) before stopping. After stopping and turning around it is heading 46 km/h (E).
\[\Delta v1=53km/h (W) - (23 km/h (W)) = 30 km/h (W)\]
After the cheetah turns around:
\[\Delta v2 = 23km/h (W) - (46 km/h (E)) = 69 km/h (W)\]
\[\Delta v =(30 + 69) km/h (W) = 99 km/h (W)\]
These are the velocities of the cheetah relative to the cameraman/
You can see the velocities and thus the accelerations are the same.

Answer 20

No, it is the object that's changing velocity -- but the acceleration is related to the force applied, and the force that the cheetah applies to the ground in order to change velocity should obviously have nothing to do with the person who's watching it happen.

Answer 21

Would you mind helping with another problem? I have posted it already.