# If a rocket with a mass of 700 kg vertically accelerates at a rate of  3/2 m/s^2, how much power will the rocket have to exert to maintain its acceleration at 4 seconds?

Jan 29, 2017

P=6.3 KW

#### Explanation:

see

MASS=700 kg
Acceleration=$\frac{3}{2} \frac{m}{s} ^ 2$

THEREFORE force exerted is $m \cdot a = 700 \cdot \frac{3}{2} = 1050 N$

Now

$a = \frac{v - u}{t}$ and we have U ,i.e initial velocity =0 as it started from rest and ofcourse every rockets before launching is at rest,lol.

so by this we can get $v$ and $t$ we are given 4 seconds

$\frac{3}{2} = \frac{v}{4}$
$v = 6 \frac{m}{s}$

therefore

$E = F \cdot D i s p l a c e m e n t$
dividing both sides of the equation by $t$

$\frac{E}{t} = F \cdot \frac{\mathrm{di} s p l a c e m e n t}{t}$

And by definition power is energy per unit time and velocity is displacement per unit time.so applying these

$p = F \cdot v$
$P = 1050 \cdot 6$ Watt

$P = 6300 W$
$P = 6.3 K W$