# An electric toy car with a mass of #2 kg# is powered by a motor with a voltage of #8 V# and a current supply of #6 A#. How long will it take for the toy car to accelerate from rest to #6 m/s#?

The product of voltage and current is the power input.

The vehicle's change in kinetic energy is

The energy divided by the power equals the amount of time required.

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To find the time it takes for the toy car to accelerate from rest to 6 m/s, we can use the equation of motion:

[ v = u + at ]

Where:

- ( v ) is the final velocity (6 m/s)
- ( u ) is the initial velocity (0 m/s, since the car starts from rest)
- ( a ) is the acceleration
- ( t ) is the time taken

We need to find the acceleration first. The force causing the acceleration can be calculated using Ohm's law:

[ V = IR ]

Where:

- ( V ) is the voltage (8 V)
- ( I ) is the current (6 A)
- ( R ) is the resistance of the motor

Given that ( V = IR ), we can rearrange it to find the resistance of the motor:

[ R = \frac{V}{I} = \frac{8, \text{V}}{6, \text{A}} = \frac{4}{3}, \Omega ]

Now, we use the formula ( F = ma ), where ( F ) is the force applied by the motor, ( m ) is the mass of the car, and ( a ) is the acceleration. The force is also given by ( F = \frac{V^2}{R} ).

Equating the two expressions for force:

[ ma = \frac{V^2}{R} ]

[ a = \frac{V^2}{mR} = \frac{(8, \text{V})^2}{(2, \text{kg})\left(\frac{4}{3}, \Omega\right)} = \frac{64}{2 \times \frac{4}{3}} , \text{m/s}^2 = 24 , \text{m/s}^2 ]

Now, using the equation of motion:

[ 6, \text{m/s} = 0 + (24 , \text{m/s}^2) t ]

[ t = \frac{6, \text{m/s}}{24, \text{m/s}^2} = \frac{1}{4}, \text{s} ]

So, it will take ( \frac{1}{4} ) seconds for the toy car to accelerate from rest to 6 m/s.

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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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- Why are high voltages used when supplying electricity over long distances?
- Two charges of # -5 C # and # -7 C# are positioned on a line at points # 5# and # 4 #, respectively. What is the net force on a charge of # -5 C# at # 0 #?

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