Physics Chapter 2 PDF

# Chapter 2 ## 7. If the velocity of a particle is nonzero, can its acceleration be zero? Explain. **Answer** Yes. If the velocity of the particle is nonzero, the particle is in motion. If the acceleration is zero, the velocity of the particle is unchanging, or is a constant. ## 8. If the veloci...

# Chapter 2 ## 7. If the velocity of a particle is nonzero, can its acceleration be zero? Explain. **Answer** Yes. If the velocity of the particle is nonzero, the particle is in motion. If the acceleration is zero, the velocity of the particle is unchanging, or is a constant. ## 8. If the velocity of a particle is zero, can its acceleration be nonzero? Explain. **Answer** Yes. If you drop a doughnut from rest (v = 0), then its acceleration is not zero. A common misconception is that immediately after the doughnut is released, both the velocity and acceleration are zero. If the acceleration were zero, then the velocity would not change, leaving the doughnut floating at rest in mid-air. ## 9. Two cars are moving in the same direction in parallel lanes along a highway. At some instant, the velocity of car A exceeds the velocity of car B. Does this mean that the acceleration of A is greater than that of B? Explain. **Answer** No: Car A might have greater acceleration than B, but they might both have zero acceleration, or otherwise equal accelerations; or the driver of B might have tramped hard on the gas pedal in the recent past. ## 13. Secretariat won the Kentucky Derby with times for successive quarter-mile segments of 25.2 s, 24.0 s, 23.8 s, and 23.0 s. (a) Find his average speed during each quarter-mile segment. (b) Assuming that Secretariat's instantaneous speed at the finish line was the same as the average speed during the final quarter mile, find his average acceleration for the entire race. (Horses in the Derby start from rest.) **(a) ** The average speed during a time interval ∆t is $\bar{v}$ = $\frac{distance traveled}{\Delta t}$. During the first quarter mile segment, Secretariat's average speed was $\bar{v_1}$ = $\frac{0.250 mi}{25.2s}$ = $\frac{1320 ft}{25.2s}$ = 52.4 ft/s (35.6 mi/h). **For the third quarter mile of the race,** $\bar{v_3}$ = $\frac{1320 ft}{23.8s}$ = 55.5 ft/s (37.7 mi/h) **(b) ** Assuming that $v_f$= $\bar{v_4}$ and recognizing that $v_I$= 0, the average acceleration during the race was $\bar{a}$= $\frac{v_f - v_i}{total elapsed time}$ = $\frac{57.4 ft/s-0}{(25.2+24.0 + 23.8 + 23.0)s}$ = 0.598 ft/s². ## 18. Draw motion diagrams for (a) an object moving to the right at constant speed, (b) an object moving to the right and speeding up at a constant rate, (c) an object moving to the right and slowing down at a constant rate, (d) an object moving to the left and speeding up at a constant rate, and (e) an object moving to the left and slowing down at a constant rate. (f) How would your drawings change if the changes in speed were not uniform; that is, if the speed were not changing at a constant rate? **ANSWER** **(a)** - → = reading order - → = velocity - → = acceleration **(b)** **(c)** **(d)** **(e)** **(f)** One way of phrasing the answer: The spacing of the successive positions would change with less regularity. Another way: The object would move with some combination of the kinds of motion shown in (a) through (e). Within one drawing, the accelerations vectors would vary in magnitude and direction. ## 3. The position versus time for a certain particle moving along the x axis is shown in Figure P2.3. Find the average velocity in the time intervals (a) 0 to 2s, (b) 0 to 4s, (c) 2 s to 4 s, (d) 4 s to 7 s, (e) 0 to 8 s. **ANSWER** **(a)** $\bar{v}$ = $\frac{\Delta x}{\Delta t}$ = $\frac{10 m}{2s}$ = 5 m/s **(b)** $\bar{v}$ = $\frac{5 m}{4s}$ = 1.2 m/s **(c)** $\bar{v}$ = $\frac{x_2-x_1}{t_2-t_1}$ = $\frac{5m-10m}{4s-2s}$ = -2.5 m/s **(d)** $\bar{v}$ = $\frac{x_2-x_1}{t_2-t_1}$ = $\frac{-5m-5m}{7s-4s}$ = -3.3 m/s **(e)** $\bar{v}$ = $\frac{x_2-x_1}{t_2-t_1}$ = $\frac{0-0}{8-0}$ = 0 m/s ## 5. A person walks first at a constant speed of 5.00 m/s along a straight line from point A to point B and then back along the line from B to A at a constant speed of 3.00 m/s. What is (a) her average speed over the entire trip? (b) her average velocity over the entire trip? **(a)** Let d represent the distance between A and B. Let $t_1$ be the time for which the walker has the higher speed in 5.00 m/s = $\frac{d}{t_1}$. Let $t_2$ represent the longer time for the return trip in -3.00 m/s = $\frac{d}{t_2}$. Then the times are $t_1$= $\frac{d}{(5.00 m/s)}$ and $t_2$= $\frac{d}{(3.00 m/s)}$. The average speed is: $\bar{v}$ = $\frac{Total distance}{Total time}$ = $\frac{d+d}{d/(5.00 m/s) + d/(3.00 m/s)}$ = $\frac{2d}{(8.00 m/s)d/(15.0 m²/s²)}$ $\bar{v}$ = $\frac{2(15.0 m²/s²)}{8.00 m/s}$ = 3.75 m/s **(b)** She starts and finishes at the same point A. With total displacement = 0, average velocity = 0. ## 6. The position of a particle moving along the x axis varies in time according to the expression x = 3t², where x is in meters and t is in seconds. Evaluate its position (a) at t = 3.00 s and (b) at 3.00 s + Δt. (c) Evaluate the limit of Δx/Δt as Δt approaches zero, to find the velocity at t = 3.00 s. **(a)** At any time, t, the position is given by x = (3.00 m/s²)t². Thus, at $t_I$= 3.00 s: $x_I$= (3.00 m/s²)(3.00 s)² = 27.0 m. **(b)** At $t_f$= 3.00 s + Δt: $x_f$= (3.00 m/s²)(3.00 s + Δt)², or $x_f$ = 27.0m+(18.0 m/s)Δt + (3.00 m/s²)(Δt)² **(c)** The instantaneous velocity at t = 3.00 s is: v= $lim_{Δt→0} \frac{x_f-x_i}{Δt}$ = $lim_{Δt→0} (18.0 m/s + (3.00 m/s²)Δt)$= [ 18.0 m/s ## 15. A particle moves along the x axis according to the equation x = 2.00 + 3.00t - 1.00t², where x is in meters and t is in seconds. At t = 3.00 s, find (a) the position of the particle, (b) its velocity, and (c) its acceleration. x = 2.00 +3.00t - t², v = $\frac{dx}{dt}$ = 3.00-2.00t, a = $\frac{dv}{dt}$ =-2.00 At t = 3.00 s: **(a)** x = (2.00 +9.00-9.00) m = 2.00 m **(b)** v = (3.00-6.00) m/s = -3.00 m/s **(c)** a = -2.00 m/s² + Problems 3,5,6,15