A biker is riding his bike at the speed of 54 km/h. As the rider approaches the circular turn on the road having radius 70 m, he reduces the speed of bike at a constant rate of 0.70 m/s2, by applying brakes. Find out the direction and magnitude of the net acceleration of the biker on this turn.
Answers
Answered by
0
ANSWER
Net acceleration is due to braking and centripetal acceleration
Due to Braking,
a
T
=0.5m/s
2
Speed of the cyclist, v=27km/h=7.5m/s
Radius of the circular turn, r=80m
Centripetal acceleration is given as:
a
c
=
r
V
2
=(7.5
2
)/80=0.70m/s
2
Since the angle between a
c
and a
T
is 900, the resultant acceleration a is given by:
a=(a
c
2
+a
T
2
)
1/2
a=(0.7
2
+0.5
2
)
1/2
=0.86m/s
2
tanθ=
a
T
a
c
where θ is the angle of the resultant with the direction of the velocity.
tanθ=
0.5
0.7
=1.4
θ=tan
−1
(1.4)=54.56
0
Similar questions