An antique car has a mass of 585 kg with its riders. It starts from rest and travels to its top speed of 18.4 m/s over a distance of 60.89 m. It drives at this speed for 31 s. It is then allowed to roll to a stop. This takes 10.7 s.
alright i found the acceleration which is 2.78 M/s^2 but i need to know How much force acts on the car at the end to make it stop and What is the coefficient of friction between the car and the pavement? but finding the coefficiant is vital for me to understand... please help.
for finding the force i have been using F=ma but it is not working. i used 585 * 2.78 which is the acceleration but the computer is telling me im wrong.
2007-12-17 01:51:46 · 5 answers · asked by molly 1 in Science & Mathematics ➔ Physics
I do not know how you got a= 2.78 m/s^2
The acceleration is change of velocity with respect to time
a=V/t= 18.4 /10.7 =1.72 m/s^2
You are correct F=ma
and the force of friction f=uN
u - coefficient of friction
N - normal force exerted by the car on the surface of motion.
Since N=mg we have
u mg= ma
u= a/g= 1.72 /9.81=0.175
2007-12-17 02:02:43 · answer #1 · answered by Edward 7 · 0⤊ 1⤋
Again, for the stopping coefficient, use the formula:
f = dv/gt (f = coefficient, dv = change in velocity, g = gravity, t=time)
For the Force, you are using the correct formula, but make sure you are in the correct units and convert if necessary.
2007-12-17 01:59:03 · answer #2 · answered by remowlms 7 · 0⤊ 1⤋
You are using the figures for the acceleration not deceleration. i.e. 18.4 m/s reduced to zero in 10.7 s
This will help you get to the answer.
2007-12-17 01:59:02 · answer #3 · answered by CTRL Freak 5 · 1⤊ 0⤋
ZOOOOOOOOOOM
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2007-12-17 01:55:24 · answer #4 · answered by Anonymous · 1⤊ 2⤋
i think you need a different formula....the momentum stuff or inertia i think.
2007-12-17 02:01:15 · answer #5 · answered by kid 1 · 0⤊ 2⤋