Question

Question 4.31 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?

Chapter Motion In A Plane Page 88

Answer 1

Given ,
Speed of cyclist = 27 km/h
= 27×5/18 = 7.5 m/s
Radius of circular track = 80m
We know,
Centripital accⁿ(ac)= v²/r
= ( 7.5)²/80 m/s²
= 225/320 m/s² = 0.7 m/s²

A/Q to question,
tangential accⁿ(at ) = 0.5m/s²

We know, anet = √(ac² + at²)
= √{(0.7)² + (0.5)²}
= √{ 0.49 + 0.25}
= √0.74 m/s²
= 0.8602 m/s²

Let anet makes ∅ with tangential direction.
Then,
tan∅ = ac/at
tan∅ = 0.7/0.5 = 7/5
∅ = tan-¹(7/5)

Answer 2

$r = 80m {s}^{ - 2} \: \\ v = 25 \times 5 \div 18 = \frac{15}{2} \: m {s}^{ - 2} \\ ac = \frac{ {v}^{2} }{r} \\ \frac{15 \times 15}{2 \times 2 \times 80} = 0.7rad {sec}^{2} \\ a = \sqrt{ {ac}^{2} } + \sqrt{ {at}^{2} } \\ = 0.86 \: m{s}^{ - 2}$