An electric toy car with a mass of #4 kg# is powered by a motor with a voltage of #7 V# and a current supply of #3 A#. How long will it take for the toy car to accelerate from rest to #5/2 m/s#?
The time is
The power of the motor is
The kinetic energy of the car is
But,
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Using the equation ( a = \frac{F}{m} ), where ( a ) is acceleration, ( F ) is force, and ( m ) is mass, and ( F = I \cdot V ) where ( I ) is current and ( V ) is voltage, we can find the acceleration. Then, using the equation ( v = u + at ), where ( v ) is final velocity, ( u ) is initial velocity (which is 0 m/s in this case), ( a ) is acceleration, and ( t ) is time, we can find the time.
First, find the force: [ F = I \cdot V = 3 , \text{A} \times 7 , \text{V} = 21 , \text{N} ]
Then, find the acceleration: [ a = \frac{F}{m} = \frac{21 , \text{N}}{4 , \text{kg}} = 5.25 , \text{m/s}^2 ]
Next, find the time: [ v = u + at ] [ 5/2 , \text{m/s} = 0 + (5.25 , \text{m/s}^2) \times t ] [ t = \frac{5/2 , \text{m/s}}{5.25 , \text{m/s}^2} = \frac{5}{2 \times 5.25} , \text{s} \approx 0.476 , \text{s} ]
So, it will take approximately 0.476 seconds for the toy car to accelerate from rest to ( \frac{5}{2} ) 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.
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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