{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

solution_pdf

# solution_pdf - a jmeri(vma367 Assignment 2 woods(000112...

This preview shows pages 1–3. Sign up to view the full content.

ajmeri (vma367) – Assignment 2 – woods – (000112) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points An electron, starting from rest and moving with a constant acceleration, travels 4 . 2 cm in 14 ms. What is the magnitude of this acceleration? Correct answer: 0 . 428571 km / s 2 . Explanation: Let : d = 4 . 2 cm = 4 . 2 × 10 5 m and t = 14 ms = 0 . 014 s . Starting from rest with constant acceleration, the distance is d = 1 2 a t 2 a = 2 d t 2 = 2 (4 . 2 × 10 5 m) (0 . 014 s) 2 1 km 1000 m = 0 . 428571 km / s 2 . 002 (part 1 of 2) 10.0 points The electrons that produce the picture in a TV set are accelerated by a very large electric force as they pass through a small region in the neck of the picture tube. This region is 1 . 1 cm in length, and the electrons enter with a speed of 1 × 10 5 m / s and leave with a speed of 1 . 5 × 10 8 m / s. What is their acceleration over this 1 . 1 cm length? Correct answer: 1 . 02273 × 10 18 m / s 2 . Explanation: Let : v f = 1 . 5 × 10 8 m / s v i = 1 × 10 5 m / s , and Δ x = 1 . 1 cm = 0 . 011 m . v 2 f = v 2 i + 2 a Δ x a = v 2 f - v 2 i x = (1 . 5 × 10 8 m / s) 2 - (1 × 10 5 m / s) 2 2 (0 . 011 m) = 1 . 02273 × 10 18 m / s 2 . 003 (part 2 of 2) 10.0 points How long is the electron in the accelerating region? Correct answer: 1 . 46569 × 10 10 s. Explanation: v f = v i + at t = v f - v i a = 1 . 5 × 10 8 m / s - 1 × 10 5 m / s 1 . 02273 × 10 18 m / s 2 = 1 . 46569 × 10 10 s . 004 10.0 points A rocket initially at rest accelerates at a rate of 60 m / s 2 for 0 . 81 min. What is its speed at the end of this time? Correct answer: 2916 m / s. Explanation: Let : a = 60 m / s 2 and t = 0 . 81 min . v f = v o + a t = a t = (60 m / s 2 ) (0 . 81 min) 60 s 1 min = 2916 m / s . 005 (part 1 of 4) 10.0 points Consider the plot below describing the veloc- ity of a car moving along a straight line with

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
ajmeri (vma367) – Assignment 2 – woods – (000112) 2 an initial position of 0 m and an initial veloc- ity of 0 m / s. 0 5 10 15 20 0 1 2 3 4 5 6 7 8 9 time (s) velocity (m/s) What is the acceleration at 7 s? Correct answer: 4 m / s 2 . Explanation: The acceleration is the slope of the velocity vs time graph from 6 s to 8 s: a 6 , 8 = v 8 - v 6 t 8 - t 6 = 20 m / s - (12 m / s) 2 s = 4 m / s 2 . 006 (part 2 of 4) 10.0 points the distance traveled during the first 5 s. Correct answer: 28 m. Explanation: For constant acceleration, the change in position is the area between the velocity curve and the time axis, or Δ x = v i Δ t + 1 2 a Δ t . For the first second, a 0 , 1 = v 1 - v 0 t 1 - t 0 = 0 m / s - (0 m / s) 1 s = 0 m / s 2 and the change in position after 1 s is x 0 , 1 = v 0 ( t 1 - t 0 ) + 1 2 a 0 , 1 ( t 1 - t 0 ) 2 = (0 m / s) (1 s) + 1 2 (0 m / s 2 ) (1 s) 2 = 0 m . From 1 s to 3 s, a 1 , 3 = v 3 - v 1 t 3 - t 1 = 8 m / s - (0 m / s) 2 s = 4 m / s 2 and the change in position is x 1 , 3 = v 1 ( t 3 - t 1 ) + 1 2 a 1 , 3 ( t 3 - t 1 ) 2 = (0 m / s) (2 s) + 1 2 (4 m / s 2 ) (2 s) 2 = 8 m . From 3 s to 4 s, a 3 , 4 = v 4 - v 3 t 4 - t 3 = 8 m / s - (8 m / s) 1 s = 0 m / s 2 and the change in position is x 3 , 4 = v 3 ( t 4 - t 3 ) + 1 2 a 3 , 4 ( t 4 - t 3 ) 2 = (8 m / s) (1 s) + 1 2 (0 m / s 2 ) (1 s) 2 = 8 m .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}