Question

A cyclist is moving with a speed of 6 \(ms^{–1}\). As he approaches a circular turn on the road of radius 120 m, he applies brakes and reduces his speed at a constant rate of 0.4 \(ms^{–2}\) . The magnitude of the net acceleration of the cyclist on the circular turn is

Answer 1

@hartnn @Mertsj @UnkleRhaukus @.Sam.

Answer 2

accelerating towards the center of the circle

Answer 3

Acceleration on an arc of a circle = \(\Large\frac{v^2}{R}\)

Answer 4

@.Sam. what would be the value of v??

Answer 5

isn't it mv^2 / r?

Answer 6

The book also showed v^2 / R , why is it so? I learnt mv^2 / R

Answer 7

\[a_c=\frac{v^2}{R}=\frac{6^2}{120}=0.3ms^{-2} \\ \\ \sum |a|=|0.3-0.4|=0.1ms^{-2}\]

Answer 8

mv^2 / r is from
\[F_c=ma_c \\ \\ F_c=m(\frac{v^2}{R})\]

Answer 9

the answer given by book is 0.5 m/s

Answer 10

Ok.

Answer 11

m/s^2 or m/s

Answer 12

oops m/s^2

Answer 13

Hmm then you add the "difference in acceleration" of 0.1m/s^2 to 0.4m/s^2 then its 0.5m/s^2