Answer:What are the statements?
Explanation:
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)
<u>Vf = 23 m/s</u>
The equation to be used is for the rectilinear motion at constant acceleration:
x = v₀t + 0.5at²
a = (v-v₀)/t
where
x is distance
v and v₀ is the final and initial velocity
t is time
a is acceleration
Because the acceleration is decelerating, that would be reported as -7.5 m/s². Substituting,
-7.5 = (0 - v₀)/t
v₀ = 7.5 t --> eqn 1
x = v₀t + 0.5at²
60 = (7.5t)(t) + 0.5(-7.5)(t²)
Solving for t,
t = 4s
Thus,
v₀ = 7.5 m/s² * 4s
v₀ = 30 m/s
