Question

An electric toy car with a mass of `3 kg` is powered by a motor with a voltage of `8 V` and a current supply of `3 A`. How long will it take for the toy car to accelerate from rest to `4 m/s`?

Answer 1

The time is 1 second.

Explanation:

Given:
`mass = 3" kg"`
`V_"motor" = 8" Volts"`
`I_"motor" = 3" Amperes"`
`Velocity = 4" m/s"`

Let t = the time of acceleration.
Let `P =` the power of motor `= (V_"motor")(I_"motor")`

The energy provided by the motor is:

`E_"motor" = Pt`

The kinetic energy of the vehicle:

`E_"kinetic" = 1/2mass(Velocity)^2`

The following step assumes no loss in the system:

`E_"motor" = E_"kinetic"`

`Pt = 1/2mass(Velocity)^2`

`t = (mass(Velocity)^2)/(2P)`

`t = (3" kg"(4" m/s")^2)/(2(8" Volts")(3" Amperes")`

`t = (3" kg"(16" m"^2"/s"^2))/(2(24" Watts")`

`t = 1" s"`

Answer 2

The time is `=1s`

Explanation:

The power is

`P=UI`

`U=8V`

`I=3A`

`P=3*8=24W`

The change in kinetic energy is

`DeltaKE=1/2mv^2-1/2m u^2`

`u=0`

`v=4ms^-1`

`m=3kg`

`DeltaKE=1/2*3*16-0`

`=24J`

But

`P=(DeltaKE)/t`

So,

`t=(DeltaKE)/P`

`=24/24=1s`