Question

A car is moving at 27 km/h. The driver applies brakes as he approaches a circular turn on the road of radius 80 m and his speed reduces at the constant rate of 0.50 m/s every second. The magnitude of net acceleration is.
A. 20 m/s
B. 250 m/s
C. 100 m/s
D. None of these

Answer 1

Since, the driver approaches a circular turn, the given velocity of the car and the radius of the road is used to find out the centripetal acceleration of the car. The rate at which the speed reduces gives the tangential acceleration. The resultant value of these two accelerations gives the net acceleration.
Formula used:
Centripetal acceleration is given by,
${{a}_{c}}=\dfrac{{{v}^{2}}}{r}$
Where, v is the velocity and r is the radius.
Net acceleration is given by,
$a=\sqrt{{{\left( {{a}_{c}} \right)}^{2}}+{{\left( {{a}_{t}} \right)}^{2}}}$
Where, ${{a}_{c}}$ is centripetal acceleration and ${{a}_{t}}$ is tangential acceleration.

Complete answer:
Given:
Speed of car = 27 km/hr
We convert the speed into m/s as follows:
$27\times \dfrac{5}{18}=7.5\text{ m/s}$
Given, radius of track = 80 m
Therefore, centripetal acceleration on the track can be calculated as:
${{a}_{c}}=\dfrac{{{v}^{2}}}{r}$
$\text{ }=\dfrac{{{\left( 7.5 \right)}^{2}}}{80}$
$\text{ }=\dfrac{56.25}{80}$
$\text{ }=0.7\text{ m/}{{\text{s}}^{2}}$

The rate at which the speed reduces is the tangential acceleration, therefore,
${{a}_{t}}=0.5\text{ m/}{{\text{s}}^{2}}$
Net acceleration is the resultant of both the accelerations and is given by,
$\text{ }a=\sqrt{{{\left( {{a}_{c}} \right)}^{2}}+{{\left( {{a}_{t}} \right)}^{2}}}$
$\text{ =}\sqrt{{{\left( 0.5 \right)}^{2}}+{{\left( 0.7 \right)}^{2}}}$
$\text{ }=\sqrt{0.25+0.49}$
$\text{ }=0.8\text{ m/}{{\text{s}}^{2}}\,$

The answer is option D.

Note:
As we know acceleration is vector quantity, hence we derive the formula for resultant vector for two vectors using the parallelogram law of vector addition, As we know that angle between tangential and Centripetal force is $90{}^\circ$, the resultant becomes
$R=\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos \theta }$
$\text{ }=\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos 90{}^\circ }$
$\text{ }=\sqrt{{{A}^{2}}+{{B}^{2}}}$