Mike's Car Stopping Force Problem: Find the Answer!

[Visual1Up]

Hey, I have the following problem:

A car going 15m/s is brought to rest in a distance of 2.0m as it strikes a mound of dirt. How large an average force is exerted by the seatbelts on a 90kg passenger as the car is stopped.

I have found one formula... F = (mass * change in velocity)/time ... and I have found stopping time to be approx (2.0/15) = 1.33 (unless I am wrong) so F = (90 (not sure about this) * 15)/1.33 = 10125N

The only problem is I have the right answer from a friend that is approx 5100N. Can someone tell me what's going wrong? Thanks!,
-Mike

 

[PhanthomJay]

Hey, I have the following problem:

A car going 15m/s is brought to rest in a distance of 2.0m as it strikes a mound of dirt. How large an average force is exerted by the seatbelts on a 90kg passenger as the car is stopped.

I have found one formula... F = (mass * change in velocity)/time ... and I have found stopping time to be approx (2.0/15) = 1.33 (unless I am wrong) it's T = d/v_avg; v_avg is not 15, what should it be?, solve for T and watch your decimal point[/color]so F = (90 (not sure about this) * 15)/1.33 = 10125N

The only problem is I have the right answer from a friend that is approx 5100N. Can someone tell me what's going wrong? Thanks!,
-Mike

see comments above.

 

[zaric]

As he had said, 15 is not the Vavg. Instead of using v to calculate, you should be using acceleration. Use the formula Vf ^ 2 = Vi ^2 + 2a DeltaX . Vf will be zero(because it comes to rest), Vi is 15, and delta X(displacement) is 2. Through the calculation, we get -56.25m/s^2 for acceleration. Force is mass times acceleration. The passenger's mass is 90kg, so we multiply the two figures, and get 5062.5. Because there's 2 significant figures, we round the number to 5100N.