# B Why doesn't v=a(t) consider a motor's RPM?

1. May 22, 2016

### david90

Given
f=ma, f/m=a
v=a(t), v=(f/m)(t)

Is it wrong to apply "v=(f/m)(t)" to situation where the force is generated by a motor? If no, then how come "v=(f/m)(t)" doesn't consider the motor's RPM? From my understanding, velocity should also be determined by the motor's rpm. For example, an electric bicycle whose motor produces 1000N @ 100 rpm (maximum) will be faster than an electric bicycle whose motor produces 1000N @ 1 rpm (maximum). However per v=(f/m)(t), the velocity of both cases are the same assuming m and t are the same.

Last edited: May 22, 2016
2. May 22, 2016

### Staff: Mentor

Hmm, maybe you meant something else. If v is velocity and a is acceleration then v=a(t) is wrong.

Edit maybe you meant v=a*t

3. May 22, 2016

### david90

yes

4. May 22, 2016

### Staff: Mentor

Assuming that (as Dale suggests) you meant $v=at$ not $v=a(t)$..... That relation only holds when $a$ is constant for the entire time you're considering (and when the initial speed is zero as well). However, $a$ is only constant if $F$ is constant, and that's not true for the motors you're describing here. Because the force produced by the motor varies with RPM, and RPM is determined by the speed, the force and the acceleration will both change with the speed so you can't use $v=at$.

5. May 22, 2016

### david90

What if we assume the motor in both cases are operating at their maximum rpm and the force at the bicycle's wheel is constant?

6. May 22, 2016

### Staff: Mentor

If they are at their maximum rpm then a=0

7. May 22, 2016

### Staff: Mentor

If the force at the bicyle's wheel remains constant, then we reach a stable equilibrium where that force is exactly equal to the air resistance slowing the bicycle down so the net force on the bicycle is zero, meaning that the acceleration is zero and the speed no longer changes.

The gearing between the motor and bicycle's wheels matters - it controls the relationship between the speed of the bicycle and the RPM of the motor. If you get the gearing exactly right, you can arrange things so that when the engine is turning at the speed at which it generates maximum force the bicycle is moving at the speed at which produces air resistance exactly equal to the output of the motor - that will be the fastest speed achievable until you install a more powerful motor.