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
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
3 answers
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
This is best explanation for a student to understand
1 answer