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

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

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Answer 1 · 2016-04-03T17:15:57

$1.4 . 2 s$ $1.4dot 2 s$

Explanation:

Electric Power $P$ $P$ is given by the expression $P = V \times I$ $P=VxxI$ , in the given problem
$= 5 \times 3 = 15 W$ $=5xx3=15W$
Mechanical Power $= Work done \div Time$ $="Work done"-:"Time"$ Also $Energy \div Time$ $"Energy"-:"Time"$
Since Work done $= \to F \cdot \to s$ $=vecFcdot vecs$ , assuming $∠ θ$ $angle theta$ between the Force and displacement vectors be $= 0$ $=0$ . Hence $cos θ = 1$ $costheta=1$ ,
$:.$ Mechanical Power $="Force"xx"Displacement"-:"Time"$
Also $"Force"="mass"xx"acceleration"$
Since Electric power = mechanical power,
$=>15="mass"xx"acceleration"xx"Displacement"xx1/"Time"$ .....(1)

For an constant acceleration of an object we know that
$v^2-u^2=2as$ .....(2)
where $v,u,a and s$ are final velocity, initial velocity, acceleration and displacement respectively.
Given $u=0$ , (2) reduces to
$v^2=2as$
$=>s=v^2/(2a)$
Inserting given values and this value of $s$ in equation (1) we obtain
$15=mxxaxxv^2/(2a)xx1/t$ , where $t$ is time taken
$15=cancel6^3xxcancelaxx((8/3)^2)/(cancel2cancel(a))xx1/t$ , solving for $t$
$t=3/15xx(8/3)^2=1.4dot 2 s$

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

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Explanation:

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An electric toy car with a mass of 6 kg6kg6 kg is powered by a motor with a voltage of 5 V5V5 V and a current supply of 3 A3A3 A. How long will it take for the toy car to accelerate from rest to 8/3 m/s83ms8/3 m/s?

1 Answer

Explanation:

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