1. A stopwatch starts while race car travels at 7 m/s from the pit area and accelerates at a uniform rate to a speed of 30 m/s in 19 s moving on a circular track of radius 511 m.

Assuming constant tangential acceleration, find
(a) the tangential acceleration, and
(b) the radial acceleration,at the instant when the speed is v = 15
Once you have both of those ... find the magnitude of a at any moment

2. A stopwatch starts while race car travels at 12.0 m/s from the pit area and accelerates at a uniform rate to a speed of 33 m/s in 16 s moving on a circular track of radius 502 m.
Assuming constant tangential acceleration, find
(a) the tangential acceleration, and
(b) the radial acceleration,at the instant when the speed is v = 17
Once you have both of those ... find the magnitude of a at any moment

3. A stopwatch starts while race car travels at 6 m/s from the pit area and accelerates at a uniform rate to a speed of 25 m/s in 3 s moving on a circular track of radius 492 m.
Assuming constant tangential acceleration, find
(a) the tangential acceleration, and
(b) the radial acceleration,at the instant when the speed is v = 6
Once you have both of those ... find the magnitude of a at any moment

To solve these problems, we need to understand the concepts of tangential acceleration and radial acceleration.

Tangential acceleration refers to the change in the magnitude of velocity along the tangent of the circular path. It is given by the formula:

(a) at = (vf - vi) / t

where vf is the final velocity, vi is the initial velocity, and t is the time taken to reach vf.

Radial acceleration refers to the acceleration towards the center of the circular path. It can be calculated using the formula:

(b) ar = v^2 / r

where v is the velocity at an instant, and r is the radius of the circular path.

Now, let's solve the first problem:

1. Given:
Initial velocity (vi) = 7 m/s
Final velocity (vf) = 30 m/s
Time taken (t) = 19 s
Radius (r) = 511 m

(a) To find the tangential acceleration, we can use the formula (a) mentioned earlier.

at = (vf - vi) / t
at = (30 - 7) / 19
at = 1.2105 m/s^2

(b) To find the radial acceleration at the instant when the speed is v = 15, we can substitute this value into the formula (b) mentioned earlier.

ar = v^2 / r
ar = 15^2 / 511
ar = 0.4418 m/s^2

To find the magnitude of acceleration at any moment, we need to combine both tangential and radial acceleration using the Pythagorean theorem:

a = sqrt(at^2 + ar^2)

2. Given:
Initial velocity (vi) = 12.0 m/s
Final velocity (vf) = 33 m/s
Time taken (t) = 16 s
Radius (r) = 502 m

(a) Tangential acceleration can be calculated using formula (a):

at = (vf - vi) / t
at = (33 - 12.0) / 16
at = 1.3125 m/s^2

(b) Radial acceleration at the instant when the speed is v = 17 can be calculated using formula (b):

ar = v^2 / r
ar = 17^2 / 502
ar = 0.5769 m/s^2

To find the magnitude of acceleration at any moment, we use the Pythagorean theorem:

a = sqrt(at^2 + ar^2)

3. Given:
Initial velocity (vi) = 6 m/s
Final velocity (vf) = 25 m/s
Time taken (t) = 3 s
Radius (r) = 492 m

(a) Tangential acceleration (at) can be calculated using formula (a):

at = (vf - vi) / t
at = (25 - 6) / 3
at = 6.33 m/s^2

(b) Radial acceleration at the instant when the speed is v = 6 can be calculated using formula (b):

ar = v^2 / r
ar = 6^2 / 492
ar = 0.0732 m/s^2

To find the magnitude of acceleration at any moment, we use the Pythagorean theorem:

a = sqrt(at^2 + ar^2)

Note: All calculations are done assuming constant tangential acceleration.