![SearchIcon](https://vmkt.vedantu.com/vmkt/PROD/png/bdcdbbd8-08a7-4688-98e6-4aa54e5e0800-1733305962725-4102606384256179.png)
A cyclist is riding with a speed of $27km/hr$ As he approaches a circular turn on the road of radius 80m, he applies brakes and reduces his speed at the constant rate of $0.5m/s$ every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn?
Answer
493.5k+ views
Hint: This question is from the concept of circular motion which is part or mechanics where the acceleration of any object or vehicle performing circular motion is divided into two perpendicular components called tangential acceleration and normal acceleration (centripetal acceleration).
Complete step-by-step answer:
Given
The speed of the cyclist is $27km/hr = 7.5m/s$
Radius of the road is 80m
Retardation due to breaking is $0.5m/{s^2}$
As we know that the centripetal acceleration is given as ${a_c} = \dfrac{{{V^2}}}{r}$
Where a is acceleration, V is the velocity and r is the radius of the circular path
Now, using the above formula to find the centripetal acceleration
${a_c} = \dfrac{{{V^2}}}{r}$
Substituting the value of r and V in the above equation, we get
$
{a_c} = \dfrac{{{{\left( {7.5} \right)}^2}}}{{80}} \\
{a_c} = 0.70m/{s^2} \\
$
As we know that the centripetal acceleration and the tangential acceleration are at 90 degree to each other, therefore the resultant acceleration is given by
$a = {\left( {a_c^2 + a_T^2} \right)^{\dfrac{1}{2}}}$
$
a = \sqrt {{{0.7}^2} + {{0.5}^2}} \\
a = 0.86m/{s^2} \\
$
Hence, the resultant acceleration of the cyclist is $0.86m/{s^2}$
The direction of the net acceleration can be found by finding the angle between resultant and the tangential acceleration which is given by
$
\tan \theta = \dfrac{{{a_c}}}{{{a_T}}} \\
\tan \theta = \dfrac{{0.7}}{{0.5}} = 1.4 \\
\theta = {54.56^0} \\
$
The direction of the net acceleration of the cyclist on the circular turn is ${54.56^0}$
Note- As any vehicle like cycle makes a turn, the force of friction acting upon the turned wheels of the car provides centripetal force required for circular motion. The acceleration is directed radially toward the center of the circle and has a magnitude equal to the square of the body's speed along the curve divided by the distance from the center of the circle to the moving body.
Complete step-by-step answer:
Given
The speed of the cyclist is $27km/hr = 7.5m/s$
Radius of the road is 80m
Retardation due to breaking is $0.5m/{s^2}$
As we know that the centripetal acceleration is given as ${a_c} = \dfrac{{{V^2}}}{r}$
Where a is acceleration, V is the velocity and r is the radius of the circular path
Now, using the above formula to find the centripetal acceleration
${a_c} = \dfrac{{{V^2}}}{r}$
Substituting the value of r and V in the above equation, we get
$
{a_c} = \dfrac{{{{\left( {7.5} \right)}^2}}}{{80}} \\
{a_c} = 0.70m/{s^2} \\
$
As we know that the centripetal acceleration and the tangential acceleration are at 90 degree to each other, therefore the resultant acceleration is given by
$a = {\left( {a_c^2 + a_T^2} \right)^{\dfrac{1}{2}}}$
$
a = \sqrt {{{0.7}^2} + {{0.5}^2}} \\
a = 0.86m/{s^2} \\
$
Hence, the resultant acceleration of the cyclist is $0.86m/{s^2}$
The direction of the net acceleration can be found by finding the angle between resultant and the tangential acceleration which is given by
$
\tan \theta = \dfrac{{{a_c}}}{{{a_T}}} \\
\tan \theta = \dfrac{{0.7}}{{0.5}} = 1.4 \\
\theta = {54.56^0} \\
$
The direction of the net acceleration of the cyclist on the circular turn is ${54.56^0}$
Note- As any vehicle like cycle makes a turn, the force of friction acting upon the turned wheels of the car provides centripetal force required for circular motion. The acceleration is directed radially toward the center of the circle and has a magnitude equal to the square of the body's speed along the curve divided by the distance from the center of the circle to the moving body.
Recently Updated Pages
Glucose when reduced with HI and red Phosphorus gives class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The highest possible oxidation states of Uranium and class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Find the value of x if the mode of the following data class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which of the following can be used in the Friedel Crafts class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
A sphere of mass 40 kg is attracted by a second sphere class 11 physics CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Statement I Reactivity of aluminium decreases when class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
10 examples of friction in our daily life
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Prokaryotic Cells and Eukaryotic Cells
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
State and prove Bernoullis theorem class 11 physics CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What organs are located on the left side of your body class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
How many valence electrons does nitrogen have class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)