Question

An electric toy car with a mass of `6 kg` is powered by a motor with a voltage of `7 V` and a current supply of `6 A`. How long will it take for the toy car to accelerate from rest to `4 m/s`?

Answer 1

`t = 8/7 ~~1.14 sec`

Explanation:

We can suggest two solutions.

Energy conservation approach leads to following solution.
The beginning kinetic energy of a toy car is zero. Ending energy is
`E = (mV^2)/2 = (6kg*(4m/(sec^2))^2)/2 = 48J` (joules)

This is exactly the amount of work (`A`) the electric motor has performed: `E=A`.

Power of the electric motor, that is its voltage times current (`P=U*I`), time (`t`) and work (`A`) are related as
`A = P*t`

Therefore,
`E = P*t = U*I*t`
and
`t = E/(U*I) = (48J)/(7V*6A) = 48/42=8/7 ~~ 1.14 sec`

Different approach is related to analyzing the force and distance.
Assume, the time to accelerate from rest (`V_0=0`) to `V=4m/sec` is `t sec`.
That means, acceleration equals to
`a = (V-V_0)/t = 4/t m/(sec^2)`

The distance covered by a toy car equals to
`S = V_0+(a*t^2)/2 = 4/t * t^2/2 = 2t` (meters)

The force of the electric motor that accelerates a toy car equals to its mass times acceleration (Second Newton's Law)
`F = m*a = 6 kg * 4/t m/(sec^2) = 24/t (kg*m)/(sec^2) = 24/t N` (newton)

The work performed by a motor equals to force times distance:
`A = F*S = 24/t*2t=48J` (joules)

The power of the electric motor `P` (amount of work it does per unit of time) equals to a product of its voltage by current (amperage):
`P = U*I = 7V*6A=42W` (watt or joule-per-second)

The work electric motor performs during the time `t` equals to its power multiplied by time:
`A = P*t`

Therefore, `48 = 42*t` from which follows
`t = 48/42=8/7 ~~ 1.14 sec`