Question

A rocket is launched from rest with a time‑dependent acceleration a ( t ) = ( p − r t ) , a(t)=(p−rt), where p = 19.00 m/s 2 p=19.00 m/s2 and r = 4.00 m/s 3 . r=4.00 m/s3. Find the maximum velocity of the rocket.

Answer 1

Answer:

45.125 m/s

Explanation:

Acceleration is given by

$a(t)=p-rt$

$p=19\ m/s^2$

r = 4 m/s

$r=4\ m/s^3$

$v(t)=\int a(t)dt=\int (p-rt)dt=pt-r\dfrac{t^2}{2}$

For maximum velocity

$0=\dfrac{d}{dt}(pt-r\dfrac{t^2}{2})\\\Rightarrow p-rt=0\\\Rightarrow t=\dfrac{p}{r}\\\Rightarrow t=\dfrac{19}{4}\\\Rightarrow t=4.75\ s$

Maximum velocity

$v_m=pt-r\dfrac{t^2}{2}\\\Rightarrow v_m=19\times 4.75-4\times\dfrac{4.75^2}{2}\\\Rightarrow v_m=45.125\ m/s$

The maximum velocity of the rocket is 45.125 m/s