A girl is riding her bicycle. When she gets to a corner, she stops to get a drin

Question

A girl is riding her bicycle. When she gets to a corner, she stops to get a drink from her water bottle. At the time, a friend passes by her, traveling at a constant speed of 8.7 m/s.
a) After 10 s, the girl gets back on her bike and travels with a constant acceleration of 2.2 m/s^2.
How long does it take for her to catch up with her friend?
b) If the girl had been on her bike and rolling along at a speed of 1.0 m/s when her friend passes, what constant acceleration would she need to catch up with her friend in the same amount of time?
(a) t1= ------- s

NOTE) please I need help with this hard question, plus there is part b) in this question

Explanation / Answer

distance travelled by friend = 8.7 m/sec * 10 sec + 8.7 m/sec * t
where t = time after whick girl catch the friend
= 87 + 8.7 t
distance travelled by girl will be same
87 + 8.7 t = 1/2 * 2.2 * t^2
we get t = 13.68 sec

now 2nd part
if girl initial velocity = 1 m/sec
time is same ,so distance = 87 + 8.7 * 13.68 = 206.016 m
206.016 = 1 * 13.68 + 1/2 * a * (13.68)^2 [s = ut +1/2 a t^2]
hence a = 2.055 m/sec2

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