Question

A motor car is travelling at 60 m/s on a circular road of radius 1200 m. It is increasing its speed at the rate of 4 m/s2. The magnitude of acceleration of the car is
3 m/s2
5 m/s2
7 m/s2
4 m/s2​

Answer 1

Answer:

Given,

a

t

=4m/s

2

v=60m/s

r=1200m

Radial acceleration, a

r

=

r

v

2

a

r

=

1200

60×60

=3m/s

2

The acceleration of the car is

a=

(a

t

)

2

+(a

r

)

2

a=

(4)

2

+(3)

2

=

25

a=5m/s

2

The correct option is C.

Answer 2

Answer:

A motor car is traveling at 60 m/s on a circular road of a radius of 1200 m. It is increasing its speed at the rate of 4 m/s2. The magnitude of the acceleration of the car is $5\ m/s^{2}$

Explanation:

Acceleration (a) of an object is the rate of change of velocity with respect to time. For an object in a uniform circular motion, the radial acceleration ($a_r$) of that object is the acceleration along the radius, directed towards the center which can be expressed as,

$a_r = \frac{v^{2} }{r}$      

Where v is the velocity of the object and r is the radius of the circular path.

From the question,

$r = 1200\ m$

$v = 60\ m/s$

Radial acceleration of the motor car,

$a_r =\frac{60^{2} }{ 1200}} = 3\ m/s^{2}$

Increased acceleration $a_i = 4\ m/s^{2}$

Net acceleration of the car $a=\sqrt{a_{r}^{2}+a_{i}^{2}}$

$a = \sqrt{3^{2}+4^{2} } = \sqrt{25} = 5\ m/s^{2}$