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

Jul 16, 2016

$\approx 0.35 s$

Explanation:

Given

• $m \to \text{Mass of the car} = 3 k g$
• $V \to \text{The supplied voltage} = 9 V$
• $I \to \text{Current supply} = 3 A$
• $u \to \text{Initial velocity of the car} = 0$

• $v \to \text{Final velocity of the car} = \frac{5}{2} = 2.5 \frac{m}{s}$

Let

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

Now we know

• $\text{The Electrical Power supplied} \left(P\right) = I \times V$

and

• $\text{The electrical energy(E) consumed in t s} = P \times t$

$\implies E = V \times I \times t$

• $\text{The gain in kinetic energy} = \frac{1}{2} m \left({v}^{2} - {u}^{2}\right) = \frac{1}{2} m {v}^{2}$

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

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

$\implies V \times I \times t = \frac{1}{2} m {v}^{2}$

$\therefore t = \frac{1}{2} \frac{m {v}^{2}}{V \times I} = \frac{1}{2} \frac{3 \times {\left(2.5\right)}^{2}}{9 \times 3} = \frac{6.25}{18} \approx 0.35 s$