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
The speed of the motorcycle 2 when it passes motorcycle 1 is 23 m/s.
The distance traveled by motorcycle 1 is calculated as follows;
[tex]x_1 = vt\\\\x_1 = 6.5 \times 10\\\\x_1 = 65 \ m[/tex]
The distance traveled by motorcycle 2 when it passes motorcycle 1 is calculated as follows;
[tex]x_2 = x + x_1\\\\x_2 = 50 \ m \ + \ 65\ m\\\\x_2 = 115 \ m[/tex]
The constant acceleration of the motorcycle 2 is calculated as;
[tex]s = u + \frac{1}{2} at^2\\\\s = 0 + \frac{1}{2} at^2\\\\a = \frac{2s}{t^2} \\\\a = \frac{2 \times 115}{10^2} \\\\a = 2.3 \ m/s^2[/tex]
The speed of the motorcycle 2 when it passes motorcycle 1 is calculated as follows;
[tex]v_f = v_0 + at\\\\v_f = 0 + (2.3)(10)\\\\v_f = 23 \ m/s[/tex]
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