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

Question

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

Charles Aeschliman · Accepted Answer

#a =0.9375 m/s^2#

I tried this:

#V= J/C# #J= kgm^2/s^2# #A=C/s#

#(5J/C*4C/s)/(8kg*8/3m/s)=#

#(20J/s)/(64/3kgm/s)=#

#(20kgm^2/s^2*1/s)/(64/3kgm/s)=#

#(20kgm^2/s^3)/(64/3kgm/s)=#

#(20kgm^2)/s^3*(3s)/(64kgm)= 0.9375 m/s^2#

Elena Albarran · Answer

undefined To calculate the time it takes for the toy car to accelerate from rest to $ \frac{8}{3} $ m/s, we can use the equation of motion:

$$ v = u + at $$

Where:
- $ v $ is the final velocity ($ \frac{8}{3} $ m/s),
- $ u $ is the initial velocity (0 m/s, since the car starts from rest),
- $ a $ is the acceleration, and
- $ t $ is the time.

The acceleration ($ a $) can be calculated using Newton's second law of motion, which states that the force ($ F $) acting on an object is equal to the mass ($ m $) of the object multiplied by its acceleration ($ a $):

$$ F = ma $$

We already know the force ($ F $) acting on the toy car, which is the product of the voltage ($ V $) and the current ($ I $) supplied to the motor:

$$ F = VI $$

Substituting the given values:

$$ F = 5 \, \text{V} \times 4 \, \text{A} = 20 \, \text{N} $$

Now, we can rearrange Newton's second law to solve for acceleration ($ a $):

$$ a = \frac{F}{m} = \frac{20 \, \text{N}}{8 \, \text{kg}} = 2.5 \, \text{m/s}^2 $$

Now, we can use the equation of motion to find the time ($ t $):

$$ \frac{8}{3} = 0 + (2.5)t $$

Solving for $ t $:

$$ t = \frac{\frac{8}{3}}{2.5} \approx 1.067 \, \text{s} $$

So, it will take approximately 1.067 seconds for the toy car to accelerate from rest to $ \frac{8}{3} $ m/s.