A car starts from rest and travels along a circular track with a radius of 55.0 meters. The car starts out at a position due East of the center of the track, heading due North initially around the track, which is counterclockwise. The car increases its speed by 105 m/s every minute around the track. What is the cars acceleration towards the center of the track when it is 1/4 of the way around the track?

v^2=u^2+2as

v^2=2as
2pi*r/4=86.35

v^2=302.225

centripetal acceleration=v^2/r=5.49

To find the car's acceleration towards the center of the track when it is 1/4 of the way around the track, we need to calculate the car's instantaneous acceleration at that point.

The car is initially at rest, and its speed increases by 105 m/s every minute around the track. This means that after t minutes, the car's speed will be 105t m/s.

To determine the car's acceleration, we can use the formula for centripetal acceleration:

a = v^2 / r

where a is the acceleration, v is the velocity or speed, and r is the radius of the circular track.

Since the car is traveling counterclockwise, the velocity vector will always be tangent to the circular path. At any given point, the velocity vector is perpendicular to the radius vector. In this case, when the car is at 1/4 of the way around the track, the velocity vector is pointing due West, and the radius vector is pointing due South towards the center of the track.

The magnitude of the velocity vector can be found using the Pythagorean theorem:

v = √(105t)^2 + 55^2

At 1/4 of the way around the track, the car has traveled a quarter of the total distance around the track. The circumference of a circle is given by 2πr, so the total distance around the track is:

d = 2π(55.0)

Since the car is 1/4 of the way around the track, it has traveled:

d/4 = (2π(55.0))/4

Now, we can substitute the values into the acceleration formula:

a = (√(105t)^2 + 55^2) / 55.0

Therefore, the formula to calculate the car's acceleration towards the center of the track when it is 1/4 of the way around the track is:

a = (√(105t)^2 + 55^2) / 55.0

This formula gives the magnitude of the acceleration. To determine the direction, we note that the velocity vector is perpendicular to the radius vector, so the acceleration vector points towards the center of the track, which is radially inward.