An electric toy car with a mass of $6 k g$ $6 kg$ is powered by a motor with a voltage of $4 V$ $4 V$ and a current supply of $2 A$ $2 A$ . How long will it take for the toy car to accelerate from rest to $\frac{14}{3} \frac{m}{s}$ $14/3 m/s$ ?

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Answer 1 · 2016-03-20T02:44:55

$8.17 s$ $8.17 "s"$

The power input is the product of the voltage and the current.

$P = I V$ $P = IV$

$= (4 V) \cdot (2 A)$ $= (4 "V") * (2 "A")$

$= 8 W$ $= 8 "W"$

The change in kinetic energy for the car is

$Delta "KE" = 1/2 m (v^2 - v_0^2)$

$= 1/2 (6 "kg") ((14/3 "m/s")^2 - (0 "m/s")^2)$

$= 65.3 "J"$

Thus, the time needed is the energy divided by the power.

$t = frac{Delta "KE"}{P}$

$= frac{65.3 "J"}{8 "W"}$

$= 8.17 "s"$

An electric toy car with a mass of 6 kg6kg6 kg is powered by a motor with a voltage of 4 V4V4 V and a current supply of 2 A2A2 A. How long will it take for the toy car to accelerate from rest to 14/3 m/s143ms14/3 m/s?