How Do You Calculate Force and Acceleration in a Crash Test Scenario?

[JB83]

I can not work out how to do the first part of this problem. I know that I am supposed to use impulse to find the average force, but F*dt= J. and J= mvf-mvo. I don't know how to work this problem out without using time. The distance part of the equation is also throwing me off. I don't know where to start. Any help would be greatly appreciated.


A set of crash tests consists of running a test car moving at a speed of 10.0 m/s (22.0 m/h) into a solid wall. Strapped securely in an advanced seat belt system, a 65.0 kg (143.0 lbs) dummy is found to move a distance of 0.630 m from the moment the car touches the wall to the time the car is stopped. Calculate the size of the average force which acts on the dummy during that time.

Using the direction of motion as positive direction, calculate the average acceleration of the dummy during that time (in g's) (use 1g=9.8 m/s2).

In a different car, the distance the dummy moves while being stopped is reduced from 0.630 m to 0.210 m calculate the average force on the dummy as that car stops.

 

[PhanthomJay]

I can not work out how to do the first part of this problem. I know that I am supposed to use impulse to find the average force, but F*dt= J. and J= mvf-mvo. I don't know how to work this problem out without using time. The distance part of the equation is also throwing me off. I don't know where to start. Any help would be greatly appreciated.


A set of crash tests consists of running a test car moving at a speed of 10.0 m/s (22.0 m/h) into a solid wall. Strapped securely in an advanced seat belt system, a 65.0 kg (143.0 lbs) dummy is found to move a distance of 0.630 m from the moment the car touches the wall to the time the car is stopped. Calculate the size of the average force which acts on the dummy during that time.

Using the direction of motion as positive direction, calculate the average acceleration of the dummy during that time (in g's) (use 1g=9.8 m/s2).

In a different car, the distance the dummy moves while being stopped is reduced from 0.630 m to 0.210 m calculate the average force on the dummy as that car stops.

You don't have the time, but you have the distance and the average speed during the collision. delta t = distance/avg speed. Solve for the average collision force. Then its Newton 2 to solve for a.

 

[JB83]

so if delta t = distance/ave speed... delta t = 0.630/10. Using that in the J= ave. Force * delta t would be ave Force = J/delta t. If J= mvf-mvo. Since there is no final velocity J = -mvo. When I did these equations, the answer that I got was incorrect.

 

[PhanthomJay]

so if delta t = distance/ave speed... delta t = 0.630/10. Using that in the J= ave. Force * delta t would be ave Force = J/delta t. If J= mvf-mvo. Since there is no final velocity J = -mvo. When I did these equations, the answer that I got was incorrect.

That's on account of you did not compute the average speed correctly. Starts at 10. Ends up at 0. Average speeed is ?

 

[JB83]

Ok I feel somewhat silly, and I know you are trying your best to explain it to me, but I have no idea what you are trying to lead me to do. Average speed is distance/change in time... I don't know what to do. Thank you for your help though.

 

[PhanthomJay]

Ok I feel somewhat silly, and I know you are trying your best to explain it to me, but I have no idea what you are trying to lead me to do. Average speed is distance/change in time... I don't know what to do. Thank you for your help though.

Yes, that is correct, you are trying to find the time, given the distance and the average speed.
v_{average} = (v_i + v_f)/2 = 10/2 = 5. Now if that equation disturbs you, why not use v_0^2 = 2a\Delta x, solve for a and use F=ma to solve for the avg force. More than one way to skin a cat, you know.

 

[JB83]

Thank you so so much. I got it, and I still feel kinda dumb.. but you were right I need to not limit myself to the equations that are in the chapter that the homework is from. Thanks again