Question

A cyclist riding with a speed of $27\,Kmph$. 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 $0.50\,m{s^{ - 1}}$ every second. The net acceleration of cyclist on the circular turn is

Answer 1

Hint:Use the formula given below and substitute the given parameters in it to calculate the centripetal acceleration. From that again use the formula to calculate the magnitude of the net acceleration. Use the angle formula to calculate the direction of the net acceleration.

Useful formula:
(1) The formula of the centripetal acceleration is given by
$a = \dfrac{{{s^2}}}{r}$
Where $a$ is the centripetal acceleration, $s$ is the speed of the cycle and $r$ is the radius of the circular turn.
(2) The net acceleration is given by
${a_n} = {\left( {{a^2} + a_f^2} \right)^{\dfrac{1}{2}}}$
Where ${a_n}$ is the net acceleration and ${a_f}$ is the final acceleration during the process of applying brakes.
(3) The angle of the net acceleration with the velocity direction is given by
$\tan \,\theta = \dfrac{{{a_c}}}{{{a_f}}}$

Complete step by step solution:
Initial speed of the cycle, $s = 27\,Kmph = 7.5\,m{s^{ - 1}}$
Radius of the circular turn on the road, $r = 80\,m$
Final acceleration of the cycle, ${a_f} = 0.50\,m{s^{ - 1}}$
By using the formula of the centripetal acceleration,
$a = \dfrac{{{s^2}}}{r}$
Substituting the values in the above equation,
$a = \dfrac{{{{7.5}^2}}}{{80}}$
By simplifying the above equation,
$a = 0.70\,m{s^{ - 2}}$
From the above values of the centripetal acceleration, the net acceleration is calculated,
${a_n} = {\left( {{a^2} + a_f^2} \right)^{\dfrac{1}{2}}}$
Substituting the values of the centripetal acceleration and the final acceleration in the above equation,
${a_n} = {\left( {{{0.7}^2} + 0.5_{}^2} \right)^{\dfrac{1}{2}}}$
By simplifying,
${a_n} = 0.86\,m{s^{ - 2}}$
Hence the net acceleration obtained is $0.86\,m{s^{ - 2}}$.
The angle between the net acceleration and the velocity is obtained by
$\tan \,\theta = \dfrac{{{a_f}}}{{{a_c}}}$
Substituting the values of the centripetal and the final acceleration, $\tan \,\theta = \dfrac{{0.7}}{{0.5}}$
$\tan \theta = 1.4$
$\theta = {54.46^ \circ }$
Thus the net acceleration obtained is $0.86\,m{s^{ - 2}}$ and the direction of it is ${54.46^ \circ }$.

Note:The centripetal acceleration is denoted in this question, because the cyclist went the circular ride in which the acceleration is towards the center of the circular turn in the road. In this, the direction of the centripetal acceleration is always the right angle to that of the motion of the cycle.