Chapter 2 Problem Hints

13.(a) avg. vel. = (change in position) / elapsed time = [x(t₂) - x(t₁)] / (t₂ - t₁)
(b, c, and d) inst. vel. = v(t) = dx / dt ; evaluate for value of t
(e) at time t_m, particle is midway between positions at t = 2.00 s and t = 3.00 s. Find these positions, average them for x(t_m), then solve for t_m
(f) sketch graph

16. (a) v(t) = dx/dt ; evaluate at t = 1s
(b) positive v means moving in positive direction and vice versa
(c) speed is magnitude of velocity--no direction
(d) find value of t where v(t) = 0
(e) similar to (d)
(f) can v(t) be negative after t = 3s?

19. Area under curve is distance traveled since vt = d . Divide area into simple geometric shapes.

23. Use v = dx/dt and a = d²x/dt². Remember v is slope of x vs. t graph. If slope is changing then there is acceleration. Ask is slope increasing or decreasing?

31. v_avg = change in x / change in t. a_avg = change in v / change in t.
v = dx/dt and a = d²x/dt². Evaluate at given times.

37. Use constant a equations.

42. a_avg = change in v / change in t. To find g's. divide by 9.80

51. Let t_r be the reaction time and t_b be the braking time. Then total distance moved by the car is x = v_ot_r - v_ot_b + ½at_b². After brakes are applied, velocity is given by v = v_o+ at_b. Solve this for t_b, knowing that v = 0 when the car stops. Substitute this expression for t_binto the first equation for x. Plug in the two different values for v_o given in the problem to write two equations with two unknowns and solve for a and t_r.

60. (a) At the highest point, v = 0. Set v = 0 in free fall equation.
(b) Use y = v_ot - ½gt², solve for t when y = 0
(c) Careful here--remember you're plotting y vs. t where y = - ½gt² and v = dy/dt

68. Problem has two parts with different accelerations. Work each part separately, then combine.