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belka [17]
3 years ago
7

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=1

9.00 m/s2 and r = 4.00 m/s 3 . r=4.00 m/s3. Find the maximum velocity of the rocket.
Physics
1 answer:
Monica [59]3 years ago
7 0

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

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