(B) Calculate the average velocity during these two time in- tervals. Solution In the fi rst time interval, t t f t i t B t A 1 s. Therefore, using Equation 2.2 and the dis- placement calculated in (a), we fi nd that 2 m/s v x ( A : B ) x A : B t 2 m 1 s 8 m [ 4(3) 2(3) 2 ] [ 4(1) 2(1) 2 ] x B : D x f x i x D x B 2 m [ 4(1) 2(1) 2 ] [ 4(0) 2(0) 2 ] x A : B x f x i x B x A 30 CHAPTER 2 • Motion in One Dimension Conceptual Example 2.2 The Velocity of Different Objects Example 2.3 Average and Instantaneous Velocity Consider the following one-dimensional motions: (A) A ball thrown directly upward rises to a highest point and falls back into the thrower ’ s hand. (B) A race car starts from rest and speeds up to 100 m/s. (C) A spacecraft drifts through space at constant velocity. Are there any points in the mo- tion of these objects at which the instantaneous velocity has the same value as the average velocity over the entire mo- tion? If so, identify the point(s). Solution (A) The average velocity for the thrown ball is zero because the ball returns to the starting point; thus its displacement is zero. (Remember that average velocity is de fi ned as x / t .) There is one point at which the instanta- neous velocity is zero — at the top of the motion. (B) The car ’ s average velocity cannot be evaluated unam- biguously with the information given, but it must be some value between 0 and 100 m/s. Because the car will have every instantaneous velocity between 0 and 100 m/s at some time during the interval, there must be some instant at which the instantaneous velocity is equal to the average velocity. (C) Because the spacecraft ’ s instantaneous velocity is con- stant, its instantaneous velocity at any time and its average velocity over any time interval are the same. 10 8 6 4 2 0 – 2 – 4 0 1 2 3 4 t (s) x (m) Slope = 4 m/s Slope = – 2 m/s Figure 2.4 (Example 2.3) Position – time graph for a particle having an x coordinate that varies in time according to the expression x 4 t 2 t 2 . In the second time interval, t 2 s; therefore, These values are the same as the slopes of the lines joining these points in Figure 2.4. (C) Find the instantaneous velocity of the particle at t 2.5 s. Solution We can guess that this instantaneous velocity must be of the same order of magnitude as our previous results, that is, a few meters per second. By measuring the slope of the green line at t 2.5 s in Figure 2.4, we fi nd that 6 m/s v x 4 m/s v x ( B : D ) x B : D t 8 m 2 s 3 Simply to make it easier to read, we write the expression as x 4 t 2 t 2 rather than as x ( 4.00 m/s) t (2.00 m/s 2 ) t 2.00 . When an equation summarizes measurements, consider its coef fi - cients to have as many signi fi cant digits as other data quoted in a problem. Consider its coef fi cients to have the units required for di- mensional consistency. When we start our clocks at t 0, we usually do not mean to limit the precision to a single digit. Consider any zero value in this book to have as many signi fi cant fi gures as you need.
v SECTION 2.3 • Acceleration 31 (a) A car, modeled as a particle, moving along the x has velocity v xi at t t i and velocity v xf at t t f .
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- Acceleration, Velocity