# Velocity Time Graph to calculate resistive forces

1. Sep 3, 2010

### vf_one

Hi I'm having trouble with this problem and was wondering if anyone could help point me in the right direction.

1. The problem statement, all variables and given/known data

An experimental rocket-car of mass 1000 kg is at rest on a road that is inclined at 10 degrees to the horizontal. The driver starts the engine and releases the brakes and accelerates the rocket car down the slope for 9s. At that point the engine stops and a rear end parachute opens rapidly slowing the vehicle. (Shown in graph in attachment)

Estimate the rocket car's acceleration at t=9.1s immediately after the parachute opens
i) the net force down the slope
ii) the total resistive force (due to air resistance, friction etc)
at 9.1 s after the engine stops

2. The attempt at a solution

To find the acceleration at 9.1s, I would have to draw a tangent to that point on the graph and then find the slope of the tangent. Would this be correct?

i) to calculate the net force down the slop would this simply be
F=ma
m-mass of the rocket car, 1000kg
a- acceleration at 9.1s from the previous answer?

ii) I have no idea how to find the total resistive force. Can anyone give me any hints to start me off?

Thanks

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2. Sep 4, 2010

### Tusike

For using the tangent to find the acceleration you are correct. For i) and ii) I'm not sure if the question asks for 9.1secs after the start of this whole thing, which would make sense; or what you wrote, perhaps a typo, 9.1 secs after the engine stops, which would mean 18.1 secs after the whole experiment started. I couldn't open the attachment because of the format; but I'm assuming you mistyped and we're looking at .1second after the engine stopped. In this case you are correct for i). For ii), calculate what the net force down the slope would be due to gravity, then subtract the answer you got for i), since friction and air resistance are the reason the actual net force is less than what you just got.

3. Sep 4, 2010

### vf_one

Hey Tusike

Thanks for your help! Sorry that was a typo, it was meant to say '9.1s, just after the engine stops' so your assumption was right.

I still don't understand why you need to subtract the force found in i) by the forward force due to gravity.

The next question asks to calculate the net resistive force (due to air resistance on the parachute, friction etc) at t=20s, when a steady speed has been reached.

In this case the rocket car is travelling at 0 acceleration so the net resistive force would equal the net forward force. Since the engine is not turned on does this mean the net forward force is due to the gravity?

Last edited: Sep 4, 2010
4. Sep 4, 2010

### vf_one

I just did another question that was similar to this one except it was about skydiving and I was able to solve it. I think I understand this now. Thanks for your help!