Answer :

Now suppose the Particle is moving along x axis in positive x direction now its speed is reducing i.e. its de-acceleration or the direction of tangential acceleration (acceleration due to increase or decrease in speed) is towards negative x axis , there is another component of acceleration radial acceleration because direction of particle is also changing as it is following a circular path now direction of radial acceleration is perpendicular to the velocity of particle i.e. along Y axis

As shown in figure

Since the speed is decreasing by 0.5 m/s every second, the rate of change of speed is the tangential acceleration so the tangential acceleration is

a_{t} = -0.5 ms^{-2} i.e. 0.5 ms^{-2} opposite to the speed of cyclist along negative X axis

now radial acceleration of a particle undergoing circular motion is given by

a_{r} = v^{2}/r

where v is the magnitude of velocity or the speed of the particle,

r is the radius of the circular patch

now at the beginning of path speed of the particle is

v = 27 km/h

converting it to m/s

1 Km = 1000m and 1 hour = 60 × 60 = 3600 s

So v= 27 × 1000m/3600s = 27 × 5/18 = 7.5 m/s

Radius of the circular path

r = 80 m

so the radial acceleration is

a_{r}= 0.70 ms^{-2} radially inwards i.e. perpendicular to velocity of cyclist along positive y axis

so the magnitude of resultant acceleration or net acceleration of the cyclist is given by

i.e.

a = 0.86 ms^{-2}

i.e. the net acceleration of the cyclist is 0.86 ms^{-2} the direction is as shown in the figure above now let us find the angle made by the net acceleration of particle with radial acceleration 𝜽 which is given as

Where a_{r} is the magnitude of radial acceleration

here a_{r} = 0.7 ms^{-2}

and a_{t} is the magnitude of tangential acceleration

here a_{t} = 0.5 ms^{-2}

so we get

i.e. the total acceleration of cyclist is making an angle of 54.5^{o} with the tangential acceleration of cyclist

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