Question

A racing car spots a cyclist at a
distance of 100 m. It overtakes the
cyclist and then cyclist can observe the
racing car at a distance of 200 m. If the
car's speed is 5 times that of the cyclist
and the time lapse between the
moment the car racer spots the cyclist
and the last moment the cyclist
observes the racer is 5 sec, the speed of
the car racer is:​

Answer 1

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