Physics Fundamentals
2nd Edition
ISBN: 9780971313453
Author: Vincent P. Coletta
Publisher: PHYSICS CURRICULUM+INSTRUCT.INC.
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Chapter 2, Problem 36P
To determine
The theory for the motion represented by the graph and the sketch for the corresponding graph of
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Chapter 2 Solutions
Physics Fundamentals
Ch. 2 - Prob. 1QCh. 2 - Prob. 2QCh. 2 - Prob. 3QCh. 2 - Prob. 4QCh. 2 - Prob. 5QCh. 2 - Prob. 6QCh. 2 - Prob. 7QCh. 2 - Prob. 8QCh. 2 - Prob. 9QCh. 2 - Prob. 1P
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- A car moves with uniform acceleration on a road for n seconds covering distance d having initial velocity u and final v. Find the relation of acceleration a of nth sec.arrow_forwardA ball was kicked at an angle of 60 degrees with respect to the floor and after 3.2 seconds landed 10 meters ahead. Calculate the average velocity in x (vx), the initial velocity in y (viy), the initial velocity (vi), and the maximum height (ymax)arrow_forwardGiven that a particles motion is described by the position function x(t) = (1.64)t2 + (-0.05)t5 + (-1.52)t9 m, find its acceleration at time t = 0.05 s. Round your answer to 2 decimal places.arrow_forward
- Figure 2-42 shows a simple device for measuring your reaction time. It consists of a cardboard strip marked with a scale and two large dots. A friend holds the strip vertically, with thumb and forefinger at the dot on the right in Fig. 2-42. You then position your thumb and forefinger at the other dot (on the left in Fig. 2-42), being careful not to touch the strip. Your friend releases the strip, and you try to pinch it as soon as possible after you see it begin to fall. The mark at the place where you pinch the strip gives your reaction time. (a) How far from the lower dot should you place the 50.0 ms mark? How much higher should you place the marks for (b) 100, (c) 150, (d) 200, and (e) 250 ms? (For example, should the 100 ms marker be 2 times as far from the dot as the 50 ms marker? If so, give an answer of 2 times. Can you find any pattern in the answers?)arrow_forwardA particle's velocity is given by vlt) = -atj, where a = 0.725 m/s? is a constant. (a) Describe the particle's motion. In particular, is it speeding up, slowing down, or maintaining constant speed? speeding up slowing down O maintaining constant speed (b) Find the particle's velocity at t- 0, t= 17.5 s, and t- 6.40 min. (Express your answer in vector form.) vy-0 at t-0 m/s at t- 17.5 S m/s at t-6.40 min -278.4 m/sarrow_forwardThe position of a particle in millimeters is given by s = 27 – 12t + t, where t is in seconds. Plot the s-t and v-t relationships for the first 9 seconds. Determine the net displacement As during that interval and the total distance D traveled. By inspection of the s-t relationship, what conclusion can you reach regarding 3. the acceleration?arrow_forward
- Solve the velocity and displacement.arrow_forwardThe position of a particle that moves along the x-axis is given by: x= t3-3t2-45t (meters) The time t is in seconds. Determine the position (meters), velocity (m/s), and acceleration (m/s2) at t = 8 seconds. Show all necessary math and include all units.arrow_forwardA drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 8 steps forward and 5 steps backward, and so on. Each step is 1 m long and requires 1s. Plot the x-t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 15 m away from the start.arrow_forward
- I throw a ball straight up in the air. It travels straight up 2 meters and then straight back down where I catch it at the same height it was released from. The distance the ball has traveled is meters. The displacement of the ball is meters. (use the proper number of sig figs in your answer given the provided information).arrow_forwardThe components of vo are expressed as follows: Vinitial-x = Vocos(0) Vinitial-y = vosin(e) a) Let us first find the time it takes for the projectile to reach the maximum height. Using: Vfinal-y = Vinitial-y + ayt since the y-axis velocity of the projectile at the maximum height is Vfinal-y Then, = Vinitial-y + ayt Substituting the expression of vinitial-y and ay = -g, results to the following: t Thus, the time to reach the maximum height is tmax-height= We will use this time to the equation Yfinal - Yinitial = Vinitial-yt + (1/2)ayt2 if we use the time taken to reach the maximum height, therefore, the displacement will yield the maximum height, so hmax = Vinitial-yt + (1/2)ayt2 substituting, the vinitial-y expression above, results to the following hmax = t+ (1/2)ayt? Then, substituting the time, results to the following hmax = ( ) + (1/2)ayl Substituting ay = -g, results to hmax =( )- (1/2)g( simplifying the expression, yields hmax = x sinarrow_forwardThen, substituting the time, results to the following hmax = ( )+ (1/2)ay( Substituting ay = -g, results to hmax = ( )- (1/2)g( simplifying the expression, yields hmax = x sin b) The distance traveled by a projectile follows a uniform motion, meaning, velocity is constant from the start point until it reach the ground along the horizontal axis, so, the range R can be expressed as R= Vinitial-xt Substituting the initial velocity on the x-axis results to the following R = ( )t But, the time it takes a projectile to travel this distance is just twice of tmax-height, by substitution, we obtain the following: R = x 2 x ( Re-arranging and then applying the trigonometric identity sin(2x) = 2sin(x)cos(x) we arrive at the expression for the range R as R = sinarrow_forward
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