Question

A cyclist starts from rest and moves with
constant acceleration of 1 m/s2. A boy who is
48 m behind the cyclist starts moving with a
constant velocity of 10 m/s. After how much
time the boy meets the cyclist?
a) 8s
b) 12 s
c) both (1) and (2)
d)10 s​

Answer 1

Answer:

c

Explanation:

S=UT+1/2AT^2

FOR CYCLIST S=0+1/2 (1)T^2_____1

FOR BOY S=10 T

THE BOY IS 48 BEHIND

S = 10T - 48______2

for 1 and 2

1/2t^2= 10t -48

(t-8) (t-12) =0

t =8 and 12

FOLLOW ME PLEASE

Answer 2

c) both (1) and (2)

thank you