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

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

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Answer 1 · 2016-07-06T13:57:06

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

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
$E = \frac{m V^{2}}{2} = \frac{6 k g \cdot {(4 \frac{m}{{sec}^{2}})}^{2}}{2} = 48 J$ $E = (mV^2)/2 = (6kg*(4m/(sec^2))^2)/2 = 48J$ (joules)

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

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

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

Different approach is related to analyzing the force and distance.
Assume, the time to accelerate from rest ( $V_{0} = 0$ $V_0=0$ ) to $V = 4 \frac{m}{sec}$ $V=4m/sec$ is $t sec$ $t sec$ .
That means, acceleration equals to
$a = \frac{V - V_{0}}{t} = \frac{4}{t} \frac{m}{{sec}^{2}}$ $a = (V-V_0)/t = 4/t m/(sec^2)$

The distance covered by a toy car equals to
$S = V_{0} + \frac{a \cdot t^{2}}{2} = \frac{4}{t} \cdot \frac{t^{2}}{2} = 2 t$ $S = V_0+(a*t^2)/2 = 4/t * t^2/2 = 2t$ (meters)

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

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

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

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

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

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

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An electric toy car with a mass of 6kg6 kg is powered by a motor with a voltage of 7V7 V and a current supply of 6A6 A. How long will it take for the toy car to accelerate from rest to 4ms4 m/s?

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