Physics, 12.07.2019 06:00 garysingh7

A1000 kg weather rocket is launched straight up. the rocket motor provides a constant acceleration for 16 s, then the motor stops. the rocket altitude 20 s after launch is 5100m. you can ignore any effects of air resistance. what was the rocket's acceleration during the first 16 s

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Answer from: kaliaakh

The rocket's acceleration for the first 16 s is 27 m/s².

The rocket starts from rest and experiences uniform upward acceleration a for a time t₁ = 16 s. During this period it travels upwards a distance s₁.

Use the equation of motion

$s=ut+\frac{1}{2} t^2$

Substitute s₁ for s, 0 m/s for u, 16 s for t =t₁, and write an equation for s₁ in terms of a.

$s=ut+\frac{1}{2} t^2\\ s_1=(0 m/s)(16s)+\frac{1}{2} a(16s)^2\\ s_1=128a........(1)$

The engines are switched off after 16 s and for the next 4 s, the rocket travels upwards under the acceleration due to gravity g, which is directed down wards.

Write an expression of the velocity v of the rocket at the end of 16 s.

Use the equation of motion,

v=u+at

Substitute 0 m/s for u and 16 s for t =t₁,

$v=u+at\\ v=(0m/s)+a(16 s)\\ v=16a........(2)$

The rocket has a speed of v =16a at the beginning of its motion after its engines are switched off.

Determine the distance s₂ the rocket travels under the action of acceleration due to gravity g.

Use the equation of motion

$s=ut+\frac{1}{2} t^2$

Substitute s₂ for s, 16 a for u, 4 s for t and -9.8 m/s² for g.

$s=ut+\frac{1}{2} t^2\\ s_2=(16a)(4 s)-(9.8 m/s^2)(4 s)^2\\ s_2=(64a)-(78.4 m).......(3)$

The total distance s traveled by the rocket is given by,

s=s_1+s_2

Add equations (1) and (3) and substitute 5100 m for s.

$s=s_1+s_2\\ 5100m=128a+64a-(78.4 m)\\ 192 a = 5178.4\\ a= 26.97 m/s^2$

The acceleration of the rocket in the first 16 s is 27 m/s^2(2sf)

Answer from: Quest

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Answer from: Quest

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6 $\frac{m}{s^{2} }$

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

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