# 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 #1 A#. How long will it take for the toy car to accelerate from rest to #3 m/s#?

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First, calculate the power supplied to the motor using the formula ( P = VI ), where ( V ) is the voltage and ( I ) is the current.

Given:

- ( V = 12 , \text{V} )
- ( I = 1 , \text{A} )

[ P = 12 , \text{V} \times 1 , \text{A} = 12 , \text{W} ]

Next, use the formula for power: [ P = F \cdot v ]

Since the car starts from rest, the initial velocity ( u = 0 ).

So, the force (( F )) can be calculated using Newton's second law: [ F = ma ]

Given:

- ( m = 5 , \text{kg} )
- ( a = \frac{{v - u}}{{t}} = \frac{{3 , \text{m/s} - 0}}{{t}} = \frac{{3 , \text{m/s}}}{{t}} )

Substitute ( m ) and ( a ) into the force formula: [ F = 5 , \text{kg} \times \frac{{3 , \text{m/s}}}{{t}} = \frac{{15 , \text{N} \cdot \text{m/s}}}{{t}} ]

Since ( P = F \cdot v ): [ 12 , \text{W} = \frac{{15 , \text{N} \cdot \text{m/s}}}{{t}} \times 3 , \text{m/s} ]

Solve for ( t ): [ t = \frac{{15 , \text{N} \cdot \text{m/s}}}{{12 , \text{W}}} = \frac{{15}}{{12}} , \text{s} ]

[ t = \frac{{5}}{{4}} , \text{s} = 1.25 , \text{s} ]

Therefore, it will take the toy car approximately ( 1.25 , \text{s} ) to accelerate from rest to ( 3 , \text{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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