Question

A car is driving along a circular track of diameter d=.85km at a constant speed of v=28.5m/s. write an expression for the magnitude of the acceleration a of the car in terms of the given parameters

Answer 1

Answer: a = (2 v²)/d = 1.9 m/s²

Explanation:

In circular motion, the acceleration is given by:

a = v²/r = v²/(d/2) = (2 v²)/d

where v is the velocity and r is the radius of the circular path in which the vehicle is moving. d is the diameter of the circular path.

It is given that:

v = 28.5 m/s

r = d/2 = 0.85 km /2 = 0.425 km = 425 m

⇒ a = (28.5 m/s)²/425 m = 1.9 m/s²

Answer 2

An expression for the magnitude of the acceleration (a) of the car in terms of the given parameters is: $A_c = \frac{2V^2}{D}$

Given the following data:

• Diameter, D = 0.85 km

• Speed, V = 28.5 m/s.

To write an expression for the magnitude of the acceleration (a) of the car in terms of the given parameters:

The acceleration of an object along a circular track is referred to as centripetal acceleration.

Mathematically, the centripetal acceleration of an object is given by the formula:

$A_c = \frac{V^2}{r}$   .....equation 1

Where:

• Ac is the centripetal acceleration.

• r is the radius of the circular track.

• V is the velocity of an object.

But, $Radius, \;r = \frac{D}{2}$  .....equation 2

Substituting the eqn 2 into eqn 1, we have:

$A_c = \frac{V^2}{\frac{D}{2}}$

Simplifying further, we have:

$A_c = \frac{2V^2}{D}$

Read more: https://brainly.com/question/6082363?referrer=searchResults