An electric toy car with a mass of $8 k g$ $8 kg$ is powered by a motor with a voltage of $9 V$ $9 V$ and a current supply of $4 A$ $4 A$ . How long will it take for the toy car to accelerate from rest to $\frac{8}{3} \frac{m}{s}$ $8/3 m/s$ ?

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Answer 1 · 2017-03-05T15:20:53

$0.79 s$ $0.79s$

The kinetic energy into the car is given by

$E = \frac{1}{2} m v^{2} = \frac{1}{2} \times 8 \times {(\frac{8}{3})}^{2} = 28.4 J$ $E = 1/2mv^2 = 1/2 xx 8 xx (8/3)^2 = 28.4J$

The power from an electric circuit is given by

$P = I V$ $P = IV$

Mechanical power, meanwhile, is

$P = \frac{E}{t}$ $P = E/t$

so

$I V = \frac{E}{t}$ $IV = E/t$

The question is asking for $t$ $t$ , so

$t = \frac{E}{I V}$ $t = E/(IV)$

and, plugging in values that we know,

$t = \frac{28.4}{9 \times 4} = 0.79 s$ $t = 28.4/(9xx4) = 0.79s$

An electric toy car with a mass of 8kg8 kg is powered by a motor with a voltage of 9V9 V and a current supply of 4A4 A. How long will it take for the toy car to accelerate from rest to 83ms8/3 m/s?