Question

A rocket starts from rest and moves upward from the surface of the earth. For the first 10.0 s of its motion, the vertical acceleration of the rocket is given by ay=(2.70m/s3)t, where the +y-direction is upward.What is the speed of the rocket when it is 325 m above the surface of the earth?

Answer 1

Explanation:

The vertical acceleration of the rocket is given by :

$a_y=2.7t$

Acceleration is given by, $a=\dfrac{dv}{dt}$

$\dfrac{dv}{dt}=2.7t$

$v=\int\limits{2.7\ t.dt}$  

$v=\dfrac{2.7t^2}{2}$..............(1)

Velocity is given by, $v=\dfrac{dx}{dt}$

$\dfrac{dx}{dt}=\dfrac{2.7t^2}{2}$

$x=\int\limits{\dfrac{2.7}{2}t^2}.dt$

$x=\dfrac{2.7}{6}t^3$

From above equation, we can find the value of t at x = 325 m

$325=\dfrac{2.7}{6}t^3$

t = 8.97 s

Now put t = 8.97 s in equation (1) as :

$v=\dfrac{2.7(8.97)^2}{2}$

v = 108.62 m/s

So, the speed of the rocket when it is 325 meters above the surface of earth is 108.62 m/s. Hence, this is the required solution.

Answer 2

$a = 2.7t$ 
$v = \int\limits^t_0 {2.7t} \, dt = \frac{2.7}{2} t^2$ 
$x = \int\limits^t_0 {\frac{2.7}{2} t^2} \, dt = \frac{2.7}{6} t^3$

Solve for v with x = 325.