Answer:
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;
→ [tex]a=\frac{v-u}{t}[/tex]
→ [tex]a=\frac{0-150}{22}=-6.82[/tex] m/s²
Now we can find the distance by using the rule;
→ s = ut + [tex]\frac{1}{2}[/tex] at²
→ s = (150)(22) + [tex]\frac{1}{2}[/tex] (-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
→ [tex]t = \frac{v-u}{a}[/tex]
→ v = 150 m/s , u = 0 , a = 22.06 m/s²
Substitute these values in the rule
→ [tex]t=\frac{150-0}{22.06}=6.8[/tex] 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
