7 A cyclist starts from rest at point A and moves in a straight line with acceleration 0.5 m s−2 for a distance of 36 m. The cyclist then travels at constant speed for 25 s before slowing down, with constant deceleration, to come to rest at point B. The distance AB is 210 m. (i) Find the total time that the cyclist takes to travel from A to B. [5]
24 s after the cyclist leaves point A, a car starts from rest from point A, with constant acceleration
4 m s−2
, towards B. It is given that the car overtakes the cyclist while the cyclist is moving with
constant speed.
(ii) Find the time that it takes from when the cyclist starts until the car overtakes her. [5]
PLEASE SOLVE PART ii ASAP I HAVE HUGE EXAM TOMMORROW
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Equations used: 1. s = (1/2)(u+v)(t)
2. s = ut + (1/2)(a)(t^2)
At 24s, cyclist has travelled 12s of acceleration and 12s of constant v. So, distance travelled from A at 24s is 36m + 1/2(6+6)(12) = 108m. (Eq1 used)
At the point car overtakes cyclist, both cyclist and car have travelled the same distance from A.
So, S(cyclist) = S(car) (Eq2 used)
108 + 6t + (1/2)(0)(t^2) = 0t + (1/2)4(t^2)
108 + 6t = 2t^2
2t^2 - 6t - 108 = 0(quadratic equation)
(t + 6)(t - 9) = 0
t = -6 , 9
Since t > 0
t = 9 seconds
2. s = ut + (1/2)(a)(t^2)
At 24s, cyclist has travelled 12s of acceleration and 12s of constant v. So, distance travelled from A at 24s is 36m + 1/2(6+6)(12) = 108m. (Eq1 used)
At the point car overtakes cyclist, both cyclist and car have travelled the same distance from A.
So, S(cyclist) = S(car) (Eq2 used)
108 + 6t + (1/2)(0)(t^2) = 0t + (1/2)4(t^2)
108 + 6t = 2t^2
2t^2 - 6t - 108 = 0(quadratic equation)
(t + 6)(t - 9) = 0
t = -6 , 9
Since t > 0
t = 9 seconds
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