Question

A cyclist riding with a speed of 27 km per hour as he approaches circular turn on the road of radius a team applies brakes and uses his speed at constant rate of. 5 metre per second square every second what is the magnitude and direction of the net acceleration of a cyclist on the circular tone

Answer 1

Explanation:

this is the complete procedure for your question

Answer 2

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.