An electric toy car with a mass of #5 kg# is powered by a motor with a voltage of #12 V# and a current supply of #5 A#. How long will it take for the toy car to accelerate from rest to #3 m/s#?

Question

An electric toy car with a mass of #5 kg# is powered by a motor with a voltage of #12 V# and a current supply of #5 A#.  How long will it take for the toy car to accelerate from rest to #3 m/s#?

Nora Arzate · Accepted Answer

#0.375 "s"#

The power input is the product of the voltage and the current.

#P = IV#

#= (12 "V") * (5 "A")#

#= 60 "W"#

The change in kinetic energy for the car is

#Delta "KE" = 1/2 m (v^2 - v_0^2)#

#= 1/2 (5 "kg") ((3 "m/s")^2 - (0 "m/s")^2)#

#= 22.5 "J"#

Thus, the time needed is the energy divided by the power.

#t = frac{Delta "KE"}{P}#

#= frac{22.5 "J"}{60 "W"}#

#= 0.375 "s"#

Carson Alexander · Answer

undefined To find the time it takes for the toy car to accelerate from rest to 3 m/s, we can use the equation $ t = \frac{v_f - v_i}{a} $, where $ v_f $ is the final velocity (3 m/s), $ v_i $ is the initial velocity (0 m/s since the car starts from rest), and $ a $ is the acceleration.

We can find the acceleration using Newton's second law, $ F = ma $, where $ F $ is the force and $ m $ is the mass of the car.

First, calculate the force:
$$ F = VI = (12 \, V)(5 \, A) = 60 \, N $$

Then, calculate the acceleration:
$$ a = \frac{F}{m} = \frac{60 \, N}{5 \, kg} = 12 \, m/s^2 $$

Now, plug the values into the equation for time:
$$ t = \frac{v_f - v_i}{a} = \frac{3 \, m/s - 0 \, m/s}{12 \, m/s^2} = 0.25 \, s $$

So, it will take the toy car 0.25 seconds to accelerate from rest to 3 m/s.