A cruise started its trip 8:00. Its average speed was 100 km/h.
2 answers:
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Answer:2 ft/s
Explanation:
Given
Length of Pole is 9 ft
Length of Joe is 3 ft
Joe walks away from Pole at the rate 4 ft/s
Let Joe is x m away from Pole so its shadow length is x'
From Similar triangle concept
3x'=x+x'
x=2x'
and it is given
Differentiating
Answer:
The acceleration of rocket B is -22.24 m/s²
Explanation:
Two rockets are flying in the same direction and are side by side at the
instant their retrorockets fire
That means they started from the same point at the same time
Rocket A has an initial velocity of 5800 m/s
Rocket B has an initial velocity of 8600 m/s
After a time t both rockets are again side by side, the displacement of
each being zero
That means they are in the same position again → s = 0
The acceleration of rocket A is -15 m/s²
We need to find the acceleration for rocket B
We can find the time from the information of rocket A by using the
rule → s = u t + a t²
where s is the displacement, u is the initial velocity, a is the
acceleration, and t is the time
→ s = 0 , u = 5800 m/s , a = -15 m/s²
Substitute these values in the rule
→ 0 = 5800 t + (-15) t²
→ 0 = 5800 t - 7.5 t²
Add 7.5 t² for both sides
→ 7.5 t² = 5800 t
Divide both sides by 7.5 t
→ t = 773.3 s
The time for the both rocket to have displacement zero is 773.3 s
Now we can find the acceleration of the rocket B by using the same
rule above
→ u = 8600 m/s , t = 773.3 , s = 0
→ 0 = (8600)(773.3) + a (773.3)²
→ 0 = 6650380 + 298996.445 a
Subtract 6650380 from both sides
→ -6650380 = 298996.445 a
Divide both sides by 298996.445
→ a = -22.24 m/s²
<em>The acceleration of rocket B is -22.24 m/s²</em>