An electric toy car with a mass of $8 k g$ $8 kg$ is powered by a motor with a voltage of $5 V$ $5 V$ and a current supply of $4 A$ $4 A$ . How long will it take for the toy car to accelerate from rest to $\frac{8}{3} \frac{m}{s}$ $8/3 m/s$ ?

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Answer 1 · 2018-07-31T00:36:17

$a = 0.9375 \frac{m}{s^{2}}$ $a =0.9375 m/s^2$

I tried this:

$V = \frac{J}{C}$ $V= J/C$
$J = k g \frac{m^{2}}{s^{2}}$ $J= kgm^2/s^2$
$A = \frac{C}{s}$ $A=C/s$

$\frac{5 \frac{J}{C} \cdot 4 \frac{C}{s}}{8 k g \cdot \frac{8}{3} \frac{m}{s}} =$ $(5J/C*4C/s)/(8kg*8/3m/s)=$

$\frac{20 \frac{J}{s}}{\frac{64}{3} k g \frac{m}{s}} =$ $(20J/s)/(64/3kgm/s)=$

$\frac{20 k g \frac{m^{2}}{s^{2}} \cdot \frac{1}{s}}{\frac{64}{3} k g \frac{m}{s}} =$ $(20kgm^2/s^2*1/s)/(64/3kgm/s)=$

$\frac{20 k g \frac{m^{2}}{s^{3}}}{\frac{64}{3} k g \frac{m}{s}} =$ $(20kgm^2/s^3)/(64/3kgm/s)=$

$\frac{20 k g m^{2}}{s^{3}} \cdot \frac{3 s}{64 k g m} = 0.9375 \frac{m}{s^{2}}$ $(20kgm^2)/s^3*(3s)/(64kgm)= 0.9375 m/s^2$

An electric toy car with a mass of 8kg8 kg is powered by a motor with a voltage of 5V5 V and a current supply of 4A4 A. How long will it take for the toy car to accelerate from rest to 83ms8/3 m/s?