Physics Solution Manual for 1100 and 2101

# 4 s the motion from the end of the crossing until the

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: v + v0 = and 2x t These can be solved simultaneously to obtain the speed v when the train reaches the end of the crossing. Once v is known, Equation 2.4 can be used to find the time required for the train to reach a speed of 32 m/s. SOLUTION Adding the above equations and solving for v, we obtain 2x 2(20.0 m) v = 1 at + = 1 (1.6 m/s 2 )(2.4 s) + = 1.0 × 101 m/s 2 2 t 2.4 s The motion from the end of the crossing until the locomotive reaches a speed of 32 m/s requires a time v − v0 32 m/s − 1.0 ×101 m/s t= = = 14 s a 1.6 m/s 2 ______________________________________________________________________________ 42. REASONING Since 1 mile = 1609 m, a quarter-mile race is L = 402 m long. If a car crosses the finish line before reaching its maximum speed, then there is only one interval of constant acceleration to consider. We will first determine whether this is true by calculating the car’s displacement x1 while accelerating from rest to top speed from Equation 2.9 (v 2 ) 2 = v0 + 2ax , with v0 = 0 m/s and v = vmax: 2 vmax = ( 0 m/s ) + 2ax1 2 or x1 = 2 vmax 2a (1) Chapter 2 Problems 67 If x1 > L, then the car crosses the finish line before reaching top speed, and the total time for ( its race is found from Equation 2.8 x = v0t + 1 at 2 L = ( 0 m/s ) t + 1 at 2 = 1 at 2 2 2 2 ) , with x = L and v 0 t= or 2L a = 0 m/s: (2) On the other hand, if a car reaches its maximum speed before crossing the finish line, the race divides into two intervals, each with a different constant acceleration. The displacement x1 is found as given in Equation (1), but the time t1 to reach the maximum speed is most easily found from Equation 2.4 ( v = v0 + at ) , with v0 = 0 m/s and v = vmax: vmax = 0 m/s + at1 t1 = or vmax a (3) The time t2 that elapses during the rest of the race is found by solving Equation 2.8 (x = v t + 0 1 at 2 2 ). Let x2 = L − x1 represent the displacement for this part of the race. With the aid of Equation (1), this becomes x2 = L − 2 vmax . Then, since the car is at its maximum 2a speed, the acceleration is a = 0 m/s2, and the displacement is 2 ( ) 2 x2 = vmax t2 + 1 0 m/s 2 t2 = vmax t2 2 t2 = or x2 vmax v L − max 2a = L − vmax (4) = vmax vmax 2a Using this expression for t2 and Equation (3) for t1 gives the total time for a two-part race: t = t1 + t2 = L v + − max v a 2a max vmax v L + max = 2a vmax (5) SOLUTION First, we use Equation (1) to determine whether either car finishes the race while accelerating: vmax (106 m/s ) = 511 m = 2 2a 2 11.0 m/s 2 2 Car A Car B x1 = x1 = ( 2 vmax 2a = ) ( 92.4 m/s )2 ( 2 11.6 m/s 2 ) = 368 m Therefore, car A finishes the race before reaching its maximum speed, but car B has 402 m − 368 m = 34 m to travel at its maximum speed. Equation (2) gives the time for car A to reach the finish line as 68 KINEMATICS IN ONE DIMENSION t= Car A 2 ( 402 m ) 2L = = 8.55 s 2 a 11.0 m/s Equation (5) gives the time for car B to reach the finish line as t= Car B L vmax + vmax 2a = 402 m 92.4 m/s + = 8.33 s 92.4 m/s 2 11.6 m/s 2 ( ) Car B wins the race by 8.55 s − 8.33 s = 0.22 s . 43. SSM REASONING AND SOLUTION a. Once the pebble has left the slingshot, it is subject only to the acceleration due to gravity. Since the downward direction is negative, the acceleration of the pebble is –9.80 m/s 2 . The pebble is not decelerating. Since its velocity and acceleration both point downward, the magnitude of the pebble’s velocity is increasing, not decreasing. b. The displacement y traveled by the pebble as a function of the time t can be found from Equation 2.8. Using Equation 2.8, we have y = v0t + 1 a y t 2 = (–9.0 m/s)(0.50 s) + 1 (–9.80 m/s 2 )(0.50 s)2 = –5.7 m 2 2 Thus, after 0.50 s, the pebble is 5.7 m beneath the cliff-top. ______________________________________________________________________________ 44. REASONING Because there is no effect due to air resistance, the rock is in free fall from its launch until it hits the ground, so that the acceleration of the rock is always −9.8 m/s2, assuming upward to be the positive direction. In (a), we will consider the interval beginning at launch and ending 2.0 s later. In (b), we will consider the interval beginning at launch and ending 5.0 s later. Since the displacement isn’t required, Equation 2.4 ( v = v0 + at ) suffices to solve both parts of the problem. The stone slows down as it rises, so we expect the speed in (a) to be larger than 15 m/s. The speed in (b) could be smaller than 15 m/s (the rock does not reach its maximum height) or larger than 15 m/s (the rock reaches its maximum height and falls back down below its height at the 2.0-s point). SOLUTION a. For the interval from launch to t = 2.0 s, the final velocity is v = 15 m/s, the acceleration is a = −9.8 m/s2, and the initial velocity is to be found. Solving Equation 2.4 ( v = v0 + at ) for v0 gives Chapter 2 Problems 69 2 v0 = v − at = 15 m/s − ( −9.8 m/s )(2.0 s) = 35 m/s Therefore, at launch, Speed = 35 m/s b. Now we consider the interval from launch to t = 5.0 s. The initial velocity is that found in part (a), v0...
View Full Document

## This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

Ask a homework question - tutors are online