# An electric toy car with a mass of 6 kg is powered by a motor with a voltage of 5 V and a current supply of 3 A. How long will it take for the toy car to accelerate from rest to 8/3 m/s?

Apr 3, 2016

$1.4 \dot{2} s$

#### Explanation:

Electric Power $P$ is given by the expression $P = V \times I$, in the given problem
$= 5 \times 3 = 15 W$
Mechanical Power$= \text{Work done"-:"Time}$ Also $\text{Energy"-:"Time}$
Since Work done $= \vec{F} \cdot \vec{s}$ , assuming $\angle \theta$ between the Force and displacement vectors be $= 0$. Hence $\cos \theta = 1$,
$\therefore$ Mechanical Power$= \text{Force"xx"Displacement"-:"Time}$
Also $\text{Force"="mass"xx"acceleration}$
Since Electric power = mechanical power,
$\implies 15 = \text{mass"xx"acceleration"xx"Displacement"xx1/"Time}$ .....(1)

For an constant acceleration of an object we know that
${v}^{2} - {u}^{2} = 2 a s$ .....(2)
where $v , u , a \mathmr{and} s$ are final velocity, initial velocity, acceleration and displacement respectively.
Given $u = 0$, (2) reduces to
${v}^{2} = 2 a s$
$\implies s = {v}^{2} / \left(2 a\right)$
Inserting given values and this value of $s$ in equation (1) we obtain
$15 = m \times a \times {v}^{2} / \left(2 a\right) \times \frac{1}{t}$, where $t$ is time taken
$15 = {\cancel{6}}^{3} \times \cancel{a} \times \frac{{\left(\frac{8}{3}\right)}^{2}}{\cancel{2} \cancel{a}} \times \frac{1}{t}$, solving for $t$
$t = \frac{3}{15} \times {\left(\frac{8}{3}\right)}^{2} = 1.4 \dot{2} s$