Question

A particular muscle car can accelerate from a standing start to 60 mph in 3.7 seconds.

1. What is the average acceleration of the car? Express your result in m/s^2 and "g's"

2. What is the average force exerted on the car and driver by the road?

Answer 1

1. a = (60-0)/(3.7-0) = 16.2 m/s^2
2. F_normal = mg = 1870*9.8 = 18326 N

This is what I have but for # 1 what is it mean by express in "g's"?
and for #2, is it just asking for the normal force? I'm not really sure what an average force is

Answer 2

On average, the acceleration of the car should be the change in velocity over change in time, yes?

I suppose the normal force counts as one being exerted on the car, but there's also friction.

Answer 3

Oh, G's is a unit associated with an object in free fall on Earth (its acceleration).

I can't make head or tails of the second either, unless it is asking for the normal force...

Answer 4

The combined mass of the car and driver is 1870kg. < oops forgot this part for #2

Answer 5

why would I need a g? isn't is just moving horizontally?

Answer 6

Alright, sounds like it's asking for normal force.

Also, G's is more specifically a force per kilogram of mass that should resemble free fall; G-forces are felt in acceleration regardless of cause of acceleration.

Answer 7

When acceleration is expressed in "g's" it is normalized to the acceleration due to gravity. It just gives us an easy way to express acceleration. This is similar to mach number, which expresses the speed of something relative to the speed of the sounds in that gas.

The force exerted by the road is the force associated with the act of accelerating.
\[F = {\Delta v \over \Delta t}m\]

Answer 8

so for #1, will the answer be 16.2 m/s^2 = 16.2g?
and for #2, I don't see why acceleration of the car is the force exerted by the road, isn't that the force of the car moving?

Answer 9

\[G's = 16.2/9.81\approx1.8\]

Newton's Third Law states that if the car accelerated by sending power to the tires which push against the road, then the road must push back against the car.

Answer 10

oh! friction of the tire and the ground, I see now thanks

Answer 11

I imagine that is what they mean.

Answer 12

You all have made critical error in your calculations.

The question asks for an answer in terms of \[m/s^2\], and the original velocity is given in mph, thus we must convert to m/s before equating acceleration.

\[1 m/s = 2.237 mph\] approximately, so \[60mph/2.237 \approx 26.822 m/s\].

\[avg accel = a = \Delta V/\Delta t = ( (26.822 m/s - 0) / (3.7 - 0) ) = 7.249 m/s^2\]

The acceleration due to gravity on earth, g, is approximated at \[9.8 m/s^2\], thus \[7.249/9.8 = .7397 g's.\]