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Question: A motor car is travelling at \[60m/s\] on a circular road of radius \[1200m\]. It is increasing its ...

A motor car is travelling at 60m/s60m/s on a circular road of radius 1200m1200m. It is increasing its speed at the rate of 4m/s4m/s. The acceleration of the car is:
A. 3m/s23m/{{s}^{2}}
B. 6m/s26m/{{s}^{2}}
C. 5m/s25m/{{s}^{2}}
D. 7m/s27m/{{s}^{2}}

Explanation

Solution

When a body is performing uniform circular motion then the tangential acceleration of the body is zero, but the radial acceleration of the body is non-zero.
When a body is performing accelerated circular motion then the tangential acceleration of the body is non-zero, also the radial acceleration of the body is non-zero.

Complete step by step solution:
If a body is in circular motion with velocity vv in circular path of radius RR, then the radial acceleration of the body is given as
ac=v2R{{a}_{c}}=\dfrac{{{v}^{2}}}{R}
It is given that the linear speed of the body is 60m/s60m/s in circular path of radius 1200m.1200m. The speed of the body is increasing at the rate of 4m/s24m/{{s}^{2}}
The tangential acceleration of the body is given as,
at=dvdt=4m/s2{{a}_{t}}=\dfrac{dv}{dt}=4m/{{s}^{2}}
Putting the values of linear speed and the radius of the circular path in the formula for radial acceleration, we get
ar=(60)21200m/s2 =36001200m/s2 =3m/s2\begin{aligned} & {{a}_{r}}=\dfrac{{{\left( 60 \right)}^{2}}}{1200}m/{{s}^{2}} \\\ & =\dfrac{3600}{1200}m/{{s}^{2}} \\\ & =3m/{{s}^{2}} \end{aligned}
The radial acceleration and the tangential acceleration are perpendicular to each other.
So, the angle between the radial acceleration and the tangential acceleration is θ=90\theta =90{}^\circ
As acceleration is a vector quantity. The net acceleration of the body is the resultant of the radial acceleration and the tangential acceleration,
a=(ar)2+(at)2+2(ar)(at)cosθ =(3)2+(4)2+2(3)(4)cos90 =9+16m/s =25m/s =5m/s\begin{aligned} & a=\sqrt{{{\left( {{a}_{r}} \right)}^{2}}+{{\left( {{a}_{t}} \right)}^{2}}+2\left( {{a}_{r}} \right)\left( {{a}_{t}} \right)\cos \theta } \\\ & =\sqrt{{{\left( 3 \right)}^{2}}+{{\left( 4 \right)}^{2}}+2\left( 3 \right)\left( 4 \right)\cos 90{}^\circ } \\\ & =\sqrt{9+16}m/s \\\ & =\sqrt{25}m/s \\\ & =5m/s \end{aligned}

Hence, the acceleration of the body is 5m/s.

Note: - The direction of tangential acceleration is along the tangent to the circular motion.
- The direction of radial acceleration is along the radius of the circular motion.