A cyclist is Riding with a speed of 27 km/h. As he approaches a circular turn on 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
Answered by
27
Hey buddy,
# Answer-
ar = 0.86 m/s^2
θ = 54.46°
#// Explaination -
# Given-
v = 27 km/h = 7.5 m/s
r = 80 m
at = 0.5 m/s^2
# Solution-
Here, two types of acceleration work on cyclist.
a) Centripetal acceleration-
ac = v^2/r
ac = (7.5)^2/80
ac = 56.25/80
ac = 0.703 m/s^2
Centripetal acceleration works inwards along radius of turn.
b) Tangential acceleration-
at = 0.5 m/s^2
Tangential acceleration works along tangent to circle in opposite direction.
*Resultant acceleration-
ar = √(ac^2+at^2)
ar = √(0.7^2+0.5^2)
ar = √(0.49+0.25)
ar = √0.74
ar = 0.86 m/s^2
Direction of resultant acceleration-
tanθ = ac/at
tanθ = 0.7/0.5
tanθ = 1.4
θ = 54.46°
Resultant acceleration of 0.86 m/s^2 subtending angle 54.46° with velocity works on the cyclist.
Hope this is helpful...
# Answer-
ar = 0.86 m/s^2
θ = 54.46°
#// Explaination -
# Given-
v = 27 km/h = 7.5 m/s
r = 80 m
at = 0.5 m/s^2
# Solution-
Here, two types of acceleration work on cyclist.
a) Centripetal acceleration-
ac = v^2/r
ac = (7.5)^2/80
ac = 56.25/80
ac = 0.703 m/s^2
Centripetal acceleration works inwards along radius of turn.
b) Tangential acceleration-
at = 0.5 m/s^2
Tangential acceleration works along tangent to circle in opposite direction.
*Resultant acceleration-
ar = √(ac^2+at^2)
ar = √(0.7^2+0.5^2)
ar = √(0.49+0.25)
ar = √0.74
ar = 0.86 m/s^2
Direction of resultant acceleration-
tanθ = ac/at
tanθ = 0.7/0.5
tanθ = 1.4
θ = 54.46°
Resultant acceleration of 0.86 m/s^2 subtending angle 54.46° with velocity works on the cyclist.
Hope this is helpful...
Similar questions