Question 8 (1 point) Question 8 Saved A car is on a circular race track that has

Question

Question 8 (1 point) Question 8 Saved A car is on a circular race track that has a radius of 30 m. The car starts at t=0 with a speed of 32 m/s and slows down uniformly until it comes to a stop after it has completed one circuit of the track. Determine the magnitude of the initial radial acceleration (when the car is traveling with a speed of 32 m/s). (All of the following answers are in units of m/s2) Question 8 options: this is the answer 34.1

Question 9 (1 point) Question 9 Saved What is the direction of the radial acceleration at t=0? (the "hat" is to indicate that these are supposed to be the unit vectors, elearning does not allow the proper format to display the unit vector symbol) Question 9 options:

r hat

theta hat

- theta hat

-r hat

x hat

-x hat

Question 10 (1 point) Question 10 Saved What is the magnitude of the tangential acceleration at t=0? Question 10 options:

2.72

8.70

34.1

32.0

5.43

Question 11 (1 point) Question 11 Unsaved What is the direction of the tangential acceleration at t=0? Question 11 options:

r hat

theta hat

- theta hat

-r hat

xhat

-x hat

Question 12 (1point) Question 12 Unsaved What is the magnitude of the net acceleration at t=0? (in units of m/s2) Question 12 options: 34.2

36.82

31.4

12.4

1,170

Question 13 (1 point) Question 13 Unsaved What is the direction of the net acceleration at t=0? Question 13 options: (Use the angle shown in the diagram)

85.4°

4.56°

12.53°

0.8°

pls answer the Questions from 9 to 13

Explanation / Answer

Given,

r = 30 m ; u = 32 m/s ; v = 0 ;

9)We know that

ar = u^2/r = 32^2/30 = 34.13 m/s

Hence, ar = 34.13 m/s ; r hat

10)v = u + at

a = (v - u)/t

t = d/u = 2 pi r/u = 2 x 3.14 x 30/32 = 5.89 s

a = (0 - 32)/5.89 = 5.43 m/s^2

Hence, a = 5.43 m/s^2

11)-r hat

12)a = sqrt (34.13^2 + 5.43^2) = 31.4 m/s^2

13)theta = tan^-1(34.13/5.43) = 80 deg

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