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Answer:
Vf = 23 m/s
Explanation:
First we need to find the distance covered by the motorcycle 2 when it passes motorcycle 1. Using the uniform speed equation for motorcycle 1:
s₁ = v₁t₁
where,
s₁ = distance covered by motorcycle 1 = ?
v₁ = speed of motorcycle 1 = 6.5 m/s
t₁ = time = 10 s
Therefore,
s₁ = (6.5 m/s)(10 s)
s₁ = 65 m
Now, for the distance covered by motorcycle 2 at the meeting point. Since, the motorcycle started 50 m ahead of motorcycle 2. Therefore,
s₂ = s₁ + 50 m
s₂ = 65 m + 50 m
s₂ = 115 m
Now, using second equation of motion for motorcycle 2:
s₂ = Vi t + (1/2)at²
where,
Vi = initial velocity of motorcycle 2 = 0 m/s
Therefore,
115 m = (0 m/s)(10 s) + (1/2)(a)(10 s)²
a = 230 m/100 s²
a = 2.3 m/s²
Now, using 1st equation of motion:
Vf = Vi + at
Vf = 0 m/s + (2.3 m/s²)(10 s)
Vf = 23 m/s
tectonic plates rubbing against each other.
Answer:
A. Repeat the experiment to be sure the results are valid.
At 8:00 pm, the velocity of the storm is 55 mi northeast. Assuming that the direction is exactly northeast, the angle is 45°
At 11:00 pm, the velocity is 75 mi north. The angle is 90°
In vector form
55 ∠ 45°
and
75 ∠ 90°
The magnitude and direction of the average velocity is
(55 ∠ 45° + 75 ∠ 90° ) / 3
