A cyclist is riding with a speed of 27 km/h. As he approaches a circular turn on the road of radius 80 m, he applies brakes and reduces his speed at the constant rate of 0.50 m/s every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn ?
Answers
Explanation:
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
Answer:
Net acceleration is due to braking and centripetal acceleration
Due to Braking,
aT=0.5m/s2
Speed of the cyclist, v=27km/h=7.5m/s
Radius of the circular turn, r=80m
Centripetal acceleration is given as:
ac=rV2
=(7.52)/80=0.70m/s2
Since the angle between ac and aT is 900, the resultant acceleration a is given by:
a=(ac2+aT2)1/2
a=(0.72+0.52)1/2=0.86m/s2
tanθ=aTac
where θ is the angle of the resultant with the direction of the velocity.
tanθ=0.50.7=1.4
θ=tan−1(1.4)=54.560