# An electric toy car with a mass of #6 kg# is powered by a motor with a voltage of #5 V# and a current supply of #3 A#. How long will it take for the toy car to accelerate from rest to #8/3 m/s#?

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To find the time it takes for the toy car to accelerate from rest to ( \frac{8}{3} ) m/s, you can use the formula ( t = \frac{v}{a} ), where ( v ) is the final velocity and ( a ) is the acceleration. Given that the initial velocity (( u )) is 0 m/s, the acceleration (( a )) can be calculated using the formula ( a = \frac{F}{m} ), where ( F ) is the force produced by the motor. The force (( F )) can be calculated using ( F = I \times V ), where ( I ) is the current and ( V ) is the voltage. Then, substitute the values into the equations:

[ F = 3 , \text{A} \times 5 , \text{V} = 15 , \text{N} ] [ a = \frac{15 , \text{N}}{6 , \text{kg}} = \frac{5}{2} , \text{m/s}^2 ] [ t = \frac{\frac{8}{3} , \text{m/s}}{\frac{5}{2} , \text{m/s}^2} = \frac{16}{15} , \text{s} ]

So, it will take ( \frac{16}{15} ) seconds for the toy car to accelerate from rest to ( \frac{8}{3} ) 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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