A boy is on his bicycle. Whe he gets to a corner, he stops to drink from his wat
ID: 1967449 • Letter: A
Question
A boy is on his bicycle. Whe he gets to a corner, he stops to drink from his water bottle. At that time, a friend passes by him, traveling at a constant speed of 8.9 m/s.After 10s, the boy gets back on his bike and travels with a constant acceleration of 2.0 m/s^2. How long does it take for him to catch up with his friend?
If the boy had been on his bike and rolling along at a speed of 1.4 m/s when his friend passed, what constant acceleration would he need to catch up with his friend in the same amount of time?
Explanation / Answer
Given,velocity at which his friend is travellling=8.9m/s
distance travelled by him during 10s=8.9*10=89m
at this distance boy started on a bike with acceleration=2m/s^2
let the time at which both of them meet be 't's
distance travelled by boy on his bike=ut+1/2at^2
=0+1/2(2)(t^2)
=t^2 m
hence,89+8.9t=t^2
t^2-8.9t-89=0
t=[8.9±v(8.9^2-4*1*(-89)]/2
=[8.9±v435.21]/2
=14.881s or -5.981s
since t should be positive,
he reaches his friend after a time interval of 14.881s
B)boys constant velocity=1.4m/s
when his friends crosses him with constant velocity=8.9m/s
time taken by both to meet again=14.881s
distance travelled by his friend during this time interval=8.9*14.881=132.4409m
distance travelled by boy=ut+1/2at^2
=1.4*14.881+1/2a*14.881^2
=20.8334+110.722a
20.8334+110.722a=132.4409
a=1.00799m/s^2
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.