An electric toy car with a mass of #3 kg# is powered by a motor with a voltage of #9 V# and a current supply of #5 A#. How long will it take for the toy car to accelerate from rest to #5/2 m/s#?

Answer 1

The time is #=0.21s#

Voltage is #U=9V#
Current is #I=5A#

The poweris

#P=UI=9*5=45W#
Velocity is #v=3/2ms^-1#
Mass is #m=3kg#

The kinetic energy is

#KE=1/2mv^2=1/2*3*(5/2)^2=75/8J#

But,

#E=P*t#
The time is #t=E/P=(75/8)/45=0.21s#
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Answer 2

We can use the formula (F = ma) to find the force acting on the toy car. (F) represents force, (m) is the mass, and (a) is the acceleration. Plugging in the given mass (m = 3 , \text{kg}) and solving for (F), we find (F = (3 , \text{kg})(5/2 , \text{m/s}^2)). This gives us (F = 15/2 , \text{N}).

Using the formula (V = IR), where (V) is voltage, (I) is current, and (R) is resistance, we can find the resistance of the motor. Plugging in the given values for (V) and (I), we find (9 , \text{V} = (5 , \text{A})R), which gives us (R = 9/5 , \text{ohms}).

The force acting on the toy car is equal to the force generated by the motor, which is given by (F = \frac{V^2}{R}). Plugging in the known values for (V) and (R), we find (15/2 , \text{N} = \frac{(9 , \text{V})^2}{9/5 , \Omega}). Solving this equation for (N), we find (N = 75/2 , \text{N}).

Now, we can use the formula (F = ma) again to find the acceleration. Since we know the force and mass, we can rearrange the formula to solve for (a). This gives us (a = F/m), which gives (a = (75/2 , \text{N})/(3 , \text{kg})). Solving this, we find (a = 25/2 , \text{m/s}^2).

To find the time it takes for the car to accelerate from rest to (5/2 , \text{m/s}), we can use the formula (v = at), where (v) is the final velocity, (a) is the acceleration, and (t) is the time. Rearranging the formula to solve for (t), we find (t = v/a). Plugging in the given values for (v) and (a), we find (t = (5/2 , \text{m/s})/(25/2 , \text{m/s}^2)). Solving this equation, we find (t = 1 , \text{s}). Therefore, it will take (1 , \text{second}) for the toy car to accelerate from rest to (5/2 , \text{m/s}).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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