HW7

1. Award: 6.66 out of 6.66 points You received credit for this question in a previous attempt Score: 100/100 Points 100 % A snowboarder starts from rest at the top of a double black diamond hill. As she rides down the slope, GPS coordinates are used to determine her displacement as a function of time: x = 0.25 t 3 + t 2 + 2 t , where x and t are expressed in feet and seconds, respectively. x is measured along the surface of the hill. Determine the position, velocity, and acceleration of the boarder when t = 6.5 seconds. The position, velocity, and acceleration of the boarder are 123.91 ft, 46.7 ft/s, and 11.75 ft/s 2 , respectively when t = 6.5 seconds. References Worksheet Difficulty: Easy A snowboarder starts from rest at the top of a double black diamond hill. As she rides down the slope, GPS coordinates are used to determine her displacement as a function of time: x = 0.25 t 3 + t 2 + 2 t , where x and t are expressed in feet and seconds, respectively. x is measured along the surface of the hill. Determine the position, velocity, and acceleration of the boarder when t = 6.5 seconds. The position, velocity, and acceleration of the boarder are 123.91 ± 5% ft, 46.69 ± 5% ft/s, and 11.75 ± 5% ft/s 2 , respectively when t = 6.5 seconds. Explanation:

2. Award: 6.66 out of 6.66 points You did not receive full credit for this question in a previous attempt The motion of a particle is defined by the relation x = t 3 – 12 t 2 +36 t +32, where x and t are expressed in feet and seconds, respectively. Determine the time, position, and acceleration of the particle when v = 0 ft/s. The time, the position, and the acceleration of the particle are t 1 = 2 s, x 1 = 64 ft, and a 1 = -12 ft/s 2 The time, the position, and the acceleration of the particle are t 2 = 6 s, x 2 = 32 ft, and a 2 = 12 ft/s 2 References Worksheet Difficulty: Easy The motion of a particle is defined by the relation x = t 3 – 12 t 2 +36 t +32, where x and t are expressed in feet and seconds, respectively. Determine the time, position, and acceleration of the particle when v = 0 ft/s. The time, the position, and the acceleration of the particle are t 1 = 2 ± 5% s, x 1 = 64 ± 5% ft, and a 1 = -12 ± 5% ft/s 2 The time, the position, and the acceleration of the particle are t 2 = 6 ± 5% s, x 2 = 32 ± 5% ft, and a 2 = 12 ± 5% ft/s 2 Explanation: Differentiating,

3. Award: 6.66 out of 6.66 points You received credit for this question in a previous attempt Determine the times when the velocity is zero. (Please provide the time in ascending order.) The times when the velocity is zero are 1 s and 3 s. References Worksheet Difficulty: Medium Determine the times when the velocity is zero. (Please provide the time in ascending order.) The times when the velocity is zero are 1 ± 5% s and 3 ± 5% s. Explanation: Differentiate twice. and When velocity is zero: v = 0

4. Award: 6.66 out of 6.66 points You did not receive full credit for this question in a previous attempt Determine the position, acceleration, and total distance traveled when t = 5 s. The position is 27 ft. The acceleration is 18 ft/s 2 . The total distance is 27 ft. References Worksheet Difficulty: Medium Determine the position, acceleration, and total distance traveled when t = 5 s. The position is 27 ± 5% ft. The acceleration is 18 ± 5% ft/s 2 . The total distance is 28 ± 5% ft. Explanation: Position at t = 5 s. Acceleration at t = 5 s. Position at t = 0. x 0 = 7 ft Over 0 ≤ t ≤ 1 s, x is increasing. Over 1 s ≤ t ≤ 3 s, x is decreasing. Over 3 s ≤ t ≤ 5 s, x is increasing.

5. Award: 6.66 out of 6.66 points You received credit for this question in a previous attempt Determine how fast the car was traveling immediately before the brakes were applied. (You must provide an answer before moving to the next part.) The speed of the car immediately before the brakes were applied was 98 ft/s. References Worksheet Difficulty: Medium Determine how fast the car was traveling immediately before the brakes were applied. (You must provide an answer before moving to the next part.) The speed of the car immediately before the brakes were applied was 98 ± 5% ft/s. Explanation:

6. Award: 6.66 out of 6.66 points You received credit for this question in a previous attempt Determine the time required for the car to stop. The time required for the car to stop is 6.12 s. References Worksheet Difficulty: Medium Determine the time required for the car to stop. The time required for the car to stop is 6.13 ± 5% s. Explanation: Time to stop.

7. Award: 6.66 out of 6.66 points You received credit for this question in a previous attempt Determine the velocity of the particle when x = 5 mm. The velocity of the particle is 4.76 mm/s. References Worksheet Difficulty: Medium Determine the velocity of the particle when x = 5 mm. The velocity of the particle is 4.76 ± 5% mm/s. Explanation:

8. Award: 6.66 out of 6.66 points You received credit for this question in a previous attempt Determine the time at which the velocity of the particle is 9 mm/s. The time at which the velocity of the particle is 9 mm/s is 0.172 s. References Worksheet Difficulty: Medium Determine the time at which the velocity of the particle is 9 mm/s. The time at which the velocity of the particle is 9 mm/s is 0.171 ± 5% s. Explanation: Substitute the value of k and v = 9 mm/s and solve for t.

9. Award: 6.66 out of 6.66 points You did not receive full credit for this question in a previous attempt Determine the distance the jogger has run when t = 1.1 h. The jogger has run 7.82 km. References Worksheet Difficulty: Medium Determine the distance the jogger has run when t = 1.1 h. The jogger has run 7.82 ± 5% km. Explanation:

10. Award: 6.66 out of 6.66 points You did not receive full credit for this question in a previous attempt Determine the jogger’s acceleration in m/s 2 at t = 0. The jogger’s acceleration is – 52.1 × 10 -6 m/s 2 . References Worksheet Difficulty: Medium Determine the jogger’s acceleration in m/s 2 at t = 0. The jogger’s acceleration is – 52.1 ± 5% × 10 -6 m/s 2 . Explanation: Acceleration when t = 0. When t = 0 and x = 0, v = 7.5 km/h, . a = –52.1 × 10 -6 m/s 2

NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. An airplane begins its take-off run at A with zero velocity and a constant acceleration a . It becomes airborne 32 s later at B with a take-off velocity of 272 km/h. References Section Break Difficulty: Easy

12. Award: 6.66 out of 6.66 points You did not receive full credit for this question in a previous attempt Determine the acceleration a . (You must provide an answer before moving on to the next part.) The acceleration is 2.36 m/s 2 . References Worksheet Difficulty: Easy Determine the acceleration a . (You must provide an answer before moving on to the next part.) The acceleration is 2.3611 ± 5% m/s 2 . Explanation: Since it is constant acceleration you can find the acceleration from the distance over time.

NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Automobiles A and B are traveling in adjacent highway lanes and at t = 0 have the positions and speeds shown. Know that automobile A has a constant acceleration of 1.8 ft/s 2 and that B has a constant deceleration of 1.2 ft/s 2 . References Section Break Difficulty: Medium

14. Award: 6.66 out of 6.66 points You received credit for this question in a previous attempt Determine when and where A will overtake B . (You must provide an answer before moving on to the next part.) A will overtake B after 15.05 s at x = 734 ft. References Worksheet Difficulty: Medium Determine when and where A will overtake B . (You must provide an answer before moving on to the next part.) A will overtake B after 15.05 ± 5% s at x = 734 ± 5% ft. Explanation: