36_PartUniversity Physics Solution

36_PartUniversity Physics Solution - 2-6 2.18. Chapter 2...

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

View Full Document Right Arrow Icon
2-6 Chapter 2 (c) In part (a) the speed increases so the acceleration is in the same direction as the velocity. If the velocity direction is positive, then the acceleration is positive. In part (b) the speed decreases so the acceleration is in the direction opposite to the direction of the velocity. If the velocity direction is positive then the acceleration is negative, and if the velocity direction is negative then the acceleration direction is positive. EVALUATE: The sign of the velocity and of the acceleration indicate their direction. 2.18. IDENTIFY: The average acceleration is av- x x v a t Δ = Δ . Use ( ) x vt to find x v at each t . The instantaneous acceleration is x x dv a dt = . SET UP: (0) 3.00 m/s x v = and (5.00 s) 5.50 m/s x v = . EXECUTE: (a) 2 av- 5.50 m/s 3.00 m/s 0.500 m/s 5.00 s x x v a t Δ− == = Δ (b) 33 (0.100 m/s )(2 ) (0.200 m/s ) x x dv at t dt = . At 0 t = , 0 x a = . At 5.00 s t = , 2 1.00 m/s x a = . (c) Graphs of ( ) x and ( ) x are given in Figure 2.18. EVALUATE: () x is the slope of ( ) x and increases at t increases. The average acceleration for 0 t = to 5.00 s t = equals the instantaneous acceleration at the midpoint of the time interval, 2.50 s t = , since ( ) x is a linear function of t . Figure 2.18 2.19. (a) IDENTIFY and SET UP: x v is the slope of the x versus t curve and x a is the slope of the x v versus t curve. EXECUTE: 0 t = to 5 s t = : x versus t is a parabola so x a is a constant. The curvature is positive so x a is positive. x v versus t is a straight line with positive slope. 0 0. x v = 5 s t = to 15 s t = : x versus t is a straight line so x v is constant and 0. x a = The slope of x versus t is positive so x v is positive. 15 s t = to 25 s: t = x versus t is a parabola with negative curvature, so x a is constant and negative. x v versus t is a straight line with negative slope. The velocity is zero at 20 s, positive for 15 s to 20 s, and negative for 20 s to 25 s. 25 s t = to 35 s: t = x versus t is a straight line so x v is constant and 0. x a = The slope of x versus t is negative so x v is negative. 35 s t = to 40 s: t = x versus t is a parabola with positive curvature, so x a is constant and positive. x v versus t is a straight line with positive slope. The velocity reaches zero at 40 s. t =
Background image of page 1

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

View Full DocumentRight Arrow Icon
Motion Along a Straight Line 2-7 The graphs of ( ) x vt and ( ) x at are sketched in Figure 2.19a.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

Page1 / 5

36_PartUniversity Physics Solution - 2-6 2.18. Chapter 2...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online