# An electric toy car with a mass of 6 kg is powered by a motor with a voltage of 7 V and a current supply of 6 A. How long will it take for the toy car to accelerate from rest to 4 m/s?

Jul 6, 2016

$t = \frac{8}{7} \approx 1.14 \sec$

#### Explanation:

We can suggest two solutions.

Energy conservation approach leads to following solution.
The beginning kinetic energy of a toy car is zero. Ending energy is
$E = \frac{m {V}^{2}}{2} = \frac{6 k g \cdot {\left(4 \frac{m}{{\sec}^{2}}\right)}^{2}}{2} = 48 J$ (joules)

This is exactly the amount of work ($A$) the electric motor has performed: $E = A$.

Power of the electric motor, that is its voltage times current ($P = U \cdot I$), time ($t$) and work ($A$) are related as
$A = P \cdot t$

Therefore,
$E = P \cdot t = U \cdot I \cdot t$
and
$t = \frac{E}{U \cdot I} = \frac{48 J}{7 V \cdot 6 A} = \frac{48}{42} = \frac{8}{7} \approx 1.14 \sec$

Different approach is related to analyzing the force and distance.
Assume, the time to accelerate from rest (${V}_{0} = 0$) to $V = 4 \frac{m}{\sec}$ is $t \sec$.
That means, acceleration equals to
$a = \frac{V - {V}_{0}}{t} = \frac{4}{t} \frac{m}{{\sec}^{2}}$

The distance covered by a toy car equals to
$S = {V}_{0} + \frac{a \cdot {t}^{2}}{2} = \frac{4}{t} \cdot {t}^{2} / 2 = 2 t$ (meters)

The force of the electric motor that accelerates a toy car equals to its mass times acceleration (Second Newton's Law)
$F = m \cdot a = 6 k g \cdot \frac{4}{t} \frac{m}{{\sec}^{2}} = \frac{24}{t} \frac{k g \cdot m}{{\sec}^{2}} = \frac{24}{t} N$ (newton)

The work performed by a motor equals to force times distance:
$A = F \cdot S = \frac{24}{t} \cdot 2 t = 48 J$ (joules)

The power of the electric motor $P$ (amount of work it does per unit of time) equals to a product of its voltage by current (amperage):
$P = U \cdot I = 7 V \cdot 6 A = 42 W$ (watt or joule-per-second)

The work electric motor performs during the time $t$ equals to its power multiplied by time:
$A = P \cdot t$

Therefore, $48 = 42 \cdot t$ from which follows
$t = \frac{48}{42} = \frac{8}{7} \approx 1.14 \sec$