# If a 12 kg object is constantly accelerated from 0m/s to 40 m/s over 8 s, how much power musy be applied at t=1 ?

Mar 30, 2017

$300 \text{W}$ (Watts)

#### Explanation:

The power applied to an object is defined as $P = \frac{W}{t}$, and expanding the definition of work, $P = \frac{F d}{t}$ (assuming force and displacement are in the same direction). $\frac{d}{t} = v$, so $P = F v$. $F = m a$, so $P = m a v$.

The acceleration of an object is $\frac{\vec{{v}_{f}} - \vec{{v}_{i}}}{\Delta t}$, for this object it is (40"m/s"-0"m/s")/(8"s")=5"m/s"^2.

The velocity of an object at time $t$ is $a t$, so in this case the velocity at $1 \text{s}$ is $5 \text{m/s"^2*1"s"=5"m/s}$

Finally, solving for power using $P = m a v$, we get $P = 12 \text{kg"*5"m/s"^2*5"m/s"=300"W}$
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