Question

A cyclist is riding with a speed of 18 kmh-¹. As he approaches a circular turn on the road of radius 25√2 m, he applies brakes and reduces his speed at the constant rate of 0.5 ms-¹ every second. Determine the magnitude and direction
of the net acceleration of the cyclist on the circular turn.


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Answer 1

Given:

A cyclist is riding with a speed of 18 kmh-¹. As he approaches a circular turn on the road of radius 25√2 m, he applies brakes and reduces his speed at the constant rate of 0.5 ms-¹ every second.

To Find:

Determine the magnitude and direction  of the net acceleration of the cyclist on the circular turn.

Solution:

given,

net acceleration due to breaking, $a_{T}=0.5m/s^2$

speed of the cyclist, $v=18km/h \ = 5m/s$

radius of the circular turn , $r=25\sqrt{2} m$

Since, we know that , centripetal acceleration is given as:

$a_{c}=\dfrac{V^2}{r}\\\\=(5^2)/25\sqrt{2} \ =1/\sqrt{2}\\\\=0.70m/s^2$

Since, the angle between $a_{c}$ and $a_{T}$ is 90 degree.

therefore, the resultant acceleration is given by;

$a=(a_{c}^2+a_{T}^2)^{1/2}\\\\a=(0.7^2+0.5^2)^{1/2}\\\\a=0.86m/s^2$

$tan\theta=\dfrac{a_{c}}{a_{T}}$

where, θ is the angle of the resultant with the direction of the velocity.

here, $tan\theta=\dfrac{0.7}{0.5}=1.4\\\\\theta=tan^{-1}(1.4)= 54.56^o$

Hence, 0.86m/s^2 and 54.56 degree is the magnitude and direction of the net acceleration of the cyclist on the circular turn.