## Two motorcycles travel along a straight road heading due north. At t = 0 motorcycle 1 is at x = 50.0 m and moves with a constant speed of 6.

Question

Two motorcycles travel along a straight road heading due north. At t = 0 motorcycle 1 is at x = 50.0 m and moves with a constant speed of 6.5 m/s; motorcycle 2 starts from rest at x = 0 and moves with constant acceleration. Motorcycle 2 passes motor cycle 1 at the time t = 10.0s. What is the speed of motorcycle 2 when it passes motorcycle 1?

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Physics
19 hours
2021-07-22T03:59:36+00:00
2021-07-22T03:59:36+00:00 1 Answers
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## Answers ( )

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 = v₁t₁where,

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 mNow, 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 ms₂ = 65 m + 50 ms₂ = 115 mNow, using second equation of motion for motorcycle 2:

s₂ = Vi t + (1/2)at²where,

Vi = initial velocity of motorcycle 2 = 0 m/s

Therefore,

115 m = (0 m/s)(10 s) + (1/2)(a)(10 s)²a = 230 m/100 s²a = 2.3 m/s²Now, using 1st equation of motion:

Vf = Vi + atVf = 0 m/s + (2.3 m/s²)(10 s)Vf = 23 m/s