Physics, 20.09.2019 21:00 vale4287

Arocket car accelerates uniformly from rest at and reaches maximum velocity of 150 m/s through a distance of 510 m, the boosters turn off and the rocket coasts to a stop in 22 seconds (this time is from the moment the boosters are turned off)
what is the acceleration while the boosters are on?
how far does the booster coast from the moment the boosters are turned off?
what is the average velocity of the whole trip?

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Answer from: kaayla20

The acceleration while the boosters are ON is 22.06 m/s²

The coast is 1650 m far from the moment the boosters are turned OFF

The average velocity of whole trip is 75 m/s

Explanation:

The rocket car accelerates uniformly from rest to reach the maximum

velocity of 150 m/s through a distance 510 m

We need to find the acceleration for this part of motion

The given is:

→ The initial velocity u = 0

→ The final velocity v = 150 m/s

→ The distance s = 510 m

To find the acceleration by the given lets use the rule;

→ v² = u² + 2as

→ (150)² = 0 + 2 a (510)

→ 22500 = 1020 a

Divide both sides by 1020

→ a = 22.06 m/s²

The acceleration while the boosters are ON is 22.06 m/s²

The boosters turn off and the rocket coasts to a stop in 22 seconds

That means the rocket coast decelerated to stop in 22 seconds

We need to find how far the coast from the moment the boosters are

turned off

The given is:

→ The time t = 22 seconds

→ The final velocity v = 0

→ The initial velocity u = 150 m/s

To find the distance we must find the acceleration for this part of motion

We can do that by using the rule;

→ $a=\frac{v-u}{t}$

→ $a=\frac{0-150}{22}=-6.82$ m/s²

Now we can find the distance by using the rule;

→ s = ut + $\frac{1}{2}$ at²

→ s = (150)(22) + $\frac{1}{2}$ (-6.82)(22)²

→ s = 3300 - 1650.44 = 1649.65 ≅ 1650 m

The coast is 1650 m far from the moment the boosters are turned OFF

To find the average velocity of the whole trip we must the total time

and the total distance

→ The total distance = 510 + 1650 = 2160 m

We need to find the time of the first part of motion

→ $t = \frac{v-u}{a}$

→ v = 150 m/s , u = 0 , a = 22.06 m/s²

Substitute these values in the rule

→ $t=\frac{150-0}{22.06}=6.8$ s

→ The total time = 6.8 + 22 = 28.8 seconds

Now lets find the average velocity

→ The average velocity = total distance ÷ total time

→ The average velocity = 2160 ÷ 28.8 = 75 m/s

The average velocity of whole trip is 75 m/s

Answer from: Quest

the answer would be c. motion and direction

i hope this : )

Answer from: Quest

found in the nucleus

has mass of one amu

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