Chapter 5: Exercises with Solutions

Chapter 5Quadratic Functionsx = x 0 + v 0 t + 1 2 at20 = x 0 − x + v 0 t + 1 2 at20 = x 0 − x + v 0 t + 1 2 at2This is a quadratic equation in t and we want to solve for t, so use the quadratic formulawith a = 1 2 a, b = v 0 and c = x 0 − x.t = −b ± √ b 2 − 4ac2a√−v 0 ±t =Now subtitute in the known values.v 2 0 − 4(1 2 a)(x 0 − x)2( 1 2 a)t = −v 0 ± √ v 2 0 − 2a(x 0 − x)at = −v 0 ± √ v 2 0 − 2a(x 0 − x)at = −0 ft/s ± √ (0ft/s) 2 − 2(−32 ft/s 2 )(352 ft − 0 ft)−32 ft/s 2t = ±√ 64 ft/s 2 (352 ft)−32 ft/s 2t = ±√ 22528 ft 2 /s 2−32 ft/s 2t ≈±150.09 ft/s−32 ft/s 2t = ±150.09 ft s × 1 s × s−32 ftt = ±4.69 s≈ tWe throw out the negative result because time must be positive. Thus, the object willhit the surface of the earth in t = 4.69 s.41. The initial height of the ball is x 0 = 5 m and its initial velocity is v = 100m/s. Because we are using the metric system, we need to use the well-known value ofa = −9.8 m/s 2 for the acceleration due to gravity. We must find the time t at whichthe height is x = 0 m.Version: Fall 2007