A cyclist is travelling at a velocity of 10m/s when he reaches the top of the slope, which is 80m long. There is a bend at the bottom of the slope, which it would be dangerous to go around any faster than 11 m/s. Because of gravity, if he did not pedal or brake he would accelerate down the slope at 0.1m/s^2. To go as fast as possible but still reach the bottom at a safe speed should the cyclist brake, do nothing or pedal?
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3Can you show your working? – Kyan Cheung Oct 16 '20 at 12:12
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1Assume that the optimal acceleration is $a$. What equation do you get for it? – Matti P. Oct 16 '20 at 12:15
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1So you know that the initial speed is 10m/s; and after 80m the speed should be exactly 11 m/s. Can you form an equation from these? – Matti P. Oct 16 '20 at 12:22
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U=10m/s, s=80m ,A=0.1m/s^2 ,A=v-u/t, 0.1=v-10/t, v=0.1t+10, s=1/2(u+v)t, 80=1/2(11+0.1t+10)t – Shah21 Oct 16 '20 at 12:40
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V=10.7m/s . So the cyclist will pedal as he doesnt exceed 11m/s – Shah21 Oct 16 '20 at 12:44
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1Use $v^2 =u^2 +2as$, a standard kinematics equation. – Deepak Oct 16 '20 at 12:53
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$v_t = v_0 + at$
So, $11 = 10 + 0.1t \implies t = 10$. So the time it takes to get to a speed of $11$ m/s from $10$ m/s is $10$ seconds.
Even without any acceleration, distance covered at the initial speed of $10$ m/s in $10$ seconds is $ = 100$ m $\gt 80$ m.
So if the cyclist does nothing, he would cover the distance before reaching the max safe speed of $11$ m/s. Hence it is clear that he should pedal to be as fast as possible and still being at safe speed.
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