## In a Soapbox Derby, young participants build non-motorized cars with very low-friction wheels. Cars race by rolling down a hill. Assume that the track begins with a 47-ft-long (1 m = 3.28 ft) section tilted 11o below horizontal.

### Question:

In a Soapbox Derby, young participants build non-motorized cars with very low-friction wheels. Cars race by rolling down a hill. Assume that the track begins with a 47-ft-long (1 m = 3.28 ft) section tilted 11o below horizontal.

How do you find the area of the region bounded by the polar curves r = cos ( 2 θ ) and r = sin ( 2 θ ) ?

What is the maximum possible acceleration of a car moving down this stretch of track?

If a car starts from rest and undergoes this acceleration for the full what is its final speed in m/s?

### Answer:

Speed(velocity) and Acceleration

At the point when a body is moving, at that point the adjustment in situation of the body per unit of time is named as the speed of the body though the adjustment in the speed of the body per unit of time is named as an increasing speed of the body.

**Explanation:**

Given information:

• The length of the track is: x=61ft.

• The introductory speed of the vehicle is: v0=0v.

• The point of tendency of the slope is: θ=11∘.

(A) The articulation for the movement of the vehicle when descending the slope is,

mgsinθ=ma

a=gsinθ

Here, the mass of the vehicle is mm and the speeding up of the vehicle is a.

Substitute the given qualities.

a=9.81m/s2×sin11∘

=1.87m/s2

In this manner, the most extreme speeding up of the vehicle while descending the slope is 1.87m/s2

The articulation for the last speed of the vehicle is,

v2 = v20 + 2ax

Substitute the given qualities.

v2=0+2×1.87m/s2×(61ft×1m/3.28ft)

v=√69.5548m2/s2

=8.34m/s

Along these lines, the last speed of the vehicle is 8.34m/s8.34m/s.

## Answer ( 1 )

Answer Above.