Question

a car moves in a circle at a constant speed of 50 m/s and completes one revolution in 40s determine the magnitude of acceleration of the car​

Answer 1

5π\2

Explanation:

a=v^2/r

a=w^2×r

w=2π/T

w=v/r

2π/T=v/r

2π/40=50/r

r=1000/π

a=2π/40 × 2π/40 ×1000/r

a=5π\2

Answer 2

ANSWER:

The acceleration of car will be 7.85 \mathrm{m} / \mathrm{s}^{2}.

EXPLANATION:

   As the car is circulating in constant speed, so the acceleration experienced by the car will centripetal acceleration.

$\text { Centripetal acceleration }=\frac{v^{2}}{r}$

Here v is the speed of the car and r is the radius of the circular track.

The radius of the circular track can be determined by finding the distance covered for one revolution.

So 200 m distance is covered in one revolution. As the track is a circular track, so the distance covered in 40 s with 50 m/s speed will be equal to the circumference of the circle for one revolution.

 Here r is the radius of the circular track.

$r=\frac{2000}{2 * 3.14}=318.5 \mathrm{m}$

The "centripetal acceleration" can be calculated using the formula

$\text {Centripetal Acceleration}=\frac{v^{2}}{r}$

$\text { Centripetal Acceleration }=\frac{50 * 50}{318.5}=7.85 \mathrm{m} / \mathrm{s}^{2}$

Thus, the acceleration of car will be $7.85 \mathrm{m} / \mathrm{s}^{2}$.