An electric toy car with a mass of $2 k g$ $2 kg$ is powered by a motor with a voltage of $12 V$ $12 V$ and a current supply of $4 A$ $4 A$ . How long will it take for the toy car to accelerate from rest to $\frac{4}{5} \frac{m}{s}$ $4/5 m/s$ ?

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An electric toy car with a mass of $2 k g$ $2 kg$ is powered by a motor with a voltage of $12 V$ $12 V$ and a current supply of $4 A$ $4 A$ . How long will it take for the toy car to accelerate from rest to $\frac{4}{5} \frac{m}{s}$ $4/5 m/s$ ?

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Answer 1 · 2016-03-18T05:01:38

$0.0133 s$ $0.0133 "s"$

Explanation:

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

$P = I V$ $P = IV$

$= (12 V) \cdot (4 A)$ $= (12 "V") * (4 "A")$

$= 48 W$ $= 48 "W"$

The change in kinetic energy for the car is

$Δ KE = \frac{1}{2} m (v^{2} - v_{0}^{2})$ $Delta "KE" = 1/2 m (v^2 - v_0^2)$

$= \frac{1}{2} (2 kg) ({(\frac{4}{5} m/s)}^{2} - {(0 m/s)}^{2})$ $= 1/2 (2 "kg") ((4/5 "m/s")^2 - (0 "m/s")^2)$

$= 0.64 J$ $= 0.64 "J"$

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

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

$= \frac{0.64 J}{48 W}$ $= frac{0.64 "J"}{48 "W"}$

$= 0.0133 s$ $= 0.0133 "s"$

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An electric toy car with a mass of $2 k g$ $2 kg$ is powered by a motor with a voltage of $12 V$ $12 V$ and a current supply of $4 A$ $4 A$ . How long will it take for the toy car to accelerate from rest to $\frac{4}{5} \frac{m}{s}$ $4/5 m/s$ ?

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Explanation:

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Explanation:

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An electric toy car with a mass of $2 k g$ $2 kg$ is powered by a motor with a voltage of $12 V$ $12 V$ and a current supply of $4 A$ $4 A$ . How long will it take for the toy car to accelerate from rest to $\frac{4}{5} \frac{m}{s}$ $4/5 m/s$ ?