An electric toy car with a mass of $5 k g$ $5 kg$ is powered by a motor with a voltage of $4 V$ $4 V$ and a current supply of $5 A$ $5 A$ . How long will it take for the toy car to accelerate from rest to $\frac{3}{2} \frac{m}{s}$ $3/2 m/s$ ?

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Answer 1 · 2017-10-06T17:44:40

$time = 1.125 s$ $"time" = 1.125" s"$

${Power}_{in} =$ $"Power"_"in" =$ the power supplied by the battery

${Power}_{in} = Current \times Voltage$ $"Power"_"in" = "Current" xx "Voltage"$

${Power}_{in} = (4 V) (5 A)$ $"Power"_"in" = (4" V")(5" A")$

${Power}_{in} = 20 W$ $"Power"_"in" = 20" W"$

${Power}_{out} = \frac{Kinetic Energy}{time}$ $"Power"_"out" = ("Kinetic Energy")/("time")$

${Power}_{out} = \frac{\frac{1}{2} Mass {(Speed)}^{2}}{time}$ $"Power"_"out" = (1/2 "Mass"("Speed")^2)/("time")$

${Power}_{out} = \frac{\frac{1}{2} (5 kg) {(3 m/s)}^{2}}{time}$ $"Power"_"out" = (1/2(5" kg")(3" m/s")^2)/("time")$

${Power}_{out} = \frac{22.5 J}{time}$ $"Power"_"out" = (22.5" J")/("time")$

We are not given an efficiency of the system, therefore, we must assume that:

$P o w e r_{in} = P o w e r_{out}$ $Power_"in" = Power_"out"$

$20 W = \frac{22.5 J}{time}$ $20" W"= (22.5" J")/("time")$

$time = \frac{22.5 J}{20 W}$ $"time" = (22.5" J")/(20" W")$

$time = 1.125 s$ $"time" = 1.125" s"$

An electric toy car with a mass of 5kg5 kg is powered by a motor with a voltage of 4V4 V and a current supply of 5A5 A. How long will it take for the toy car to accelerate from rest to 32ms3/2 m/s?