Asked by valerie
the speed of a cyclist reduces uniformly from 2.5 m/s to 1.0 m/s in 12 s.
(a) calculate the deceleration of the cyclist.
(b) calculate the distance travelled by the cyclist in this time
(a) calculate the deceleration of the cyclist.
(b) calculate the distance travelled by the cyclist in this time
Answers
Answered by
Kunal
Sol
Accelaration=F.V-I.V/TIME TAKEN
NOW WE CAN SAY THAT,
a=0-2.5/12
a=-2.5/12
a=0.2083
Accelaration=F.V-I.V/TIME TAKEN
NOW WE CAN SAY THAT,
a=0-2.5/12
a=-2.5/12
a=0.2083
Answered by
ANONYMOUS
a) Acceleration or deceleration = Final velocity - initial velocity / Time taken
D = 1-2.5 / 12
D= -0.125
b) d = ½×(Vi + Vf) × t
where
d = displacement = ?
Vf = final velocity = 1 m/s
Vi = initial velocity = 2.5 m/s
t = time = 12 s
so
d = ½×(2.5 + 1) × 12
d = 21 m
D = 1-2.5 / 12
D= -0.125
b) d = ½×(Vi + Vf) × t
where
d = displacement = ?
Vf = final velocity = 1 m/s
Vi = initial velocity = 2.5 m/s
t = time = 12 s
so
d = ½×(2.5 + 1) × 12
d = 21 m
Answered by
Mavis
This is best explanation for a student to understand
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