An electric toy car with a mass of $3 k g$ $3 kg$ is powered by a motor with a voltage of $9 V$ $9 V$ and a current supply of $3 A$ $3 A$ . How long will it take for the toy car to accelerate from rest to $\frac{5}{2} \frac{m}{s}$ $5/2 m/s$ ?

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Answer 1 · 2016-07-16T04:54:30

$\approx 0.35 s$ $~~0.35s$

Given

$m \to Mass of the car = 3 k g$ $m->"Mass of the car"=3kg$
$V \to The supplied voltage = 9 V$ $V->"The supplied voltage"=9V$
$I \to Current supply = 3 A$ $I->"Current supply"=3A$
$u \to Initial velocity of the car = 0$ $u->"Initial velocity of the car"=0$
$v \to Final velocity of the car = \frac{5}{2} = 2.5 \frac{m}{s}$ $v->"Final velocity of the car"=5/2=2.5m/s$

Let

$t \to Time taken to gain the final velocity = ?$ $t->"Time taken to gain the final velocity"=?$

Now we know

$The Electrical Power supplied (P) = I \times V$ $"The Electrical Power supplied"(P)=IxxV$

and

$The electrical energy(E) consumed in t s = P \times t$ $"The electrical energy(E) consumed in t s"=Pxxt$

$\Rightarrow E = V \times I \times t$ $=>E=VxxIxxt$

$The gain in kinetic energy = \frac{1}{2} m (v^{2} - u^{2}) = \frac{1}{2} m v^{2}$ $"The gain in kinetic energy"=1/2m(v^2-u^2)=1/2mv^2$

Considering that the supplied electrical energy goes to increase the KE of the car.

So
$E = \frac{1}{2} m v^{2}$ $E=1/2mv^2$

$\Rightarrow V \times I \times t = \frac{1}{2} m v^{2}$ $=>VxxIxxt=1/2mv^2$

$:.t=1/2(mv^2)/(VxxI)=1/2(3xx(2.5)^2) /(9xx3)=6.25/18~~0.35s$

An electric toy car with a mass of 3 kg3kg3 kg is powered by a motor with a voltage of 9 V9V9 V and a current supply of 3 A3A3 A. How long will it take for the toy car to accelerate from rest to 5/2 m/s52ms5/2 m/s?