# An electric toy car with a mass of 3 kg is powered by a motor with a voltage of 8 V and a current supply of 3 A. How long will it take for the toy car to accelerate from rest to 4 m/s?

Feb 22, 2017

The time is 1 second.

#### Explanation:

Given:
$m a s s = 3 \text{ kg}$
${V}_{\text{motor" = 8" Volts}}$
${I}_{\text{motor" = 3" Amperes}}$
$V e l o c i t y = 4 \text{ m/s}$

Let t = the time of acceleration.
Let $P =$ the power of motor $= \left({V}_{\text{motor")(I_"motor}}\right)$

The energy provided by the motor is:

${E}_{\text{motor}} = P t$

The kinetic energy of the vehicle:

${E}_{\text{kinetic}} = \frac{1}{2} m a s s {\left(V e l o c i t y\right)}^{2}$

The following step assumes no loss in the system:

${E}_{\text{motor" = E_"kinetic}}$

$P t = \frac{1}{2} m a s s {\left(V e l o c i t y\right)}^{2}$

$t = \frac{m a s s {\left(V e l o c i t y\right)}^{2}}{2 P}$

$t = \left(3 \text{ kg"(4" m/s")^2)/(2(8" Volts")(3" Amperes}\right)$

$t = \left(3 \text{ kg"(16" m"^2"/s"^2))/(2(24" Watts}\right)$

$t = 1 \text{ s}$

Feb 22, 2017

The time is $= 1 s$

#### Explanation:

The power is

$P = U I$

$U = 8 V$

$I = 3 A$

$P = 3 \cdot 8 = 24 W$

The change in kinetic energy is

$\Delta K E = \frac{1}{2} m {v}^{2} - \frac{1}{2} m {u}^{2}$

$u = 0$

$v = 4 m {s}^{-} 1$

$m = 3 k g$

$\Delta K E = \frac{1}{2} \cdot 3 \cdot 16 - 0$

$= 24 J$

But

$P = \frac{\Delta K E}{t}$

So,

$t = \frac{\Delta K E}{P}$

$= \frac{24}{24} = 1 s$