Physics - One Dimensional Motion

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One Dimensional Motion

Table of Contents

Introduction

One Dimensional Motion Problems

One Dimensional Motion Answers

Introduction

Stated conventions

t₁ = 0 thus t₂ = t = elapsed time

x_o = initial position

v_o = initial velocity

x = position at time t

v = velocity at time t

x - x_o
= --------- and we can derive x = x_o + t
t

        v - v_o
a = ---------- and we can derive v = v_o + at
              t

x = x_o+ v_ot + ½ at²

v² - v_o² x = x_o + --------------- or v² = v_o² + 2a(x - x_o)
2a

Falling Bodies

Ignoring air resistance, all objects experience a uniform acceleration of 9.80 m s^-2 near the Earth’s surface. This acceleration is toward the center of the Earth. Varies slightly with latitude, altitude, and density underneath the surface. At maximum height vertical velocity = 0.

Terminal velocity - maximum velocity reached when the force of air resistance is equal to the force of gravity.

Table of Contents

One Dimensional Motion Problems

1.	If you are driving 130 km/h and take 2.0 seconds to answer your cell phone, how far did you travel during this inattentive period?
2.	A horse canters away from its trainer in a straight line, moving 150. M away in 14.0 s. It then turns abruptly and gallops halfway back in 4.5 s. Calculate the average speed and average velocity.
3.	An airplane travels 2400. km at a speed of 800. km/h, and then encounters a tailwind that boosts its speed to 1000. km/h for the next 1800. km. What is the average speed and total elapsed time?
4.	At highway speeds, an Explorer is capable of accelerating 2.5 m s^-2. How long does it take to accelerate from 100. km/h to 150. km/h?
5.	A race car has a braking distance of 50.0 m at 130. km/h. What is the acceleration in m/s² and what is the g force?
6.	A world-class sprinter can burst out of the blocks to essentially top speed of 11.5 m/s in the first 15.0 m of the race. What is the average acceleration and time necessary to reach this speed?
7.	In coming to a stop, a car leaves skid marks on the highway 110. m long. Assuming a deceleration of 4.00 m/s², what was the speed of the car before braking?
8.	A woman driving her car 60. km/h approaches an intersection just as the traffic light turns yellow. The yellow light last only 3.0 s before turning red and the car is 50. m from the beginning of the intersection. The intersection is 12 m wide and the car can decelerate 5.0 m/s² and can accelerate from 60. km/h to 90. km/h in 6.0 s. Ignoring the length of the car and reaction time, should the driver try to stop?
9.	A stone is dropped from the top of a building. It is seen to hit the ground after 5.35 s. How high is the building? What was the velocity of the stone just before it hit?
10.	A parachutist jumps from an airplane. A second parachutist jumps from the airplane 2.50 s later. How far apart are the parachutist when the second parachutist has reached 22.0 m/s?
11.	A baseball is thrown up vertically with a speed of 14.0 m/s from the edge of a 200.0 m high cliff. How much time passes before it hits the bottom? What is its speed just before hitting? What is the total distance traveled by the baseball.
12.	A person who is properly constrained by an over-the-shoulder seat belt has a good chance of surviving a car collision if the deceleration does not exceed 15 g’s. Assuming uniform deceleration, how much must the front end collapse if a crash brings the car to rest from 100. km/h?
13.	A ball is thrown upward with an initial velocity of 65.0 m/s. What is the highest point reached by the ball? How long is the ball in the air?

Table of Contents

One Dimensional Motion Answers

1.	If you are driving 130 km/h and take 2.0 seconds to answer your cell phone, how far did you travel during this inattentive period?
	km 1 h 1000 m x = v_ot = 130 ------ x ---------- x ------------- x 2.0 s = 72.2 meters h 3600 s 1 km	There is no acceleration in the direction the car is moving. Pay attention to units. Short distances are not normally given in kilometers.

2.	A horse canters away from its trainer in a straight line, moving 150. M away in 14.0 s. It then turns abruptly and gallops halfway back in 4.5 s. Calculate the average speed and average velocity.
	distance = 150. m + 75 m = 225 meters forward is + and back is - displacement = 150. m - 75 m = 75 m 225 m average speed = ----------- = 12.2 m/s 18.5 s 75 m average velocity = ----------- = 4.1 m/s 18.5 s	Average speed is based on the total distance traveled. Average velocity is based on the distance from the original position to the final position. elapsed time is the total time. pay attention to significant figures. Significant figures and Significant Figure Calculations

3.	An airplane travels 2400. km at a speed of 800. km/h, and then encounters a tailwind that boosts its speed to 1000. km/h for the next 1800. km. What is the average speed and total elapsed time?
	x x = v_ot => t = ----- v_o 2400. km t₁ = ---------------- = 3.00 h 800. km h^-1 1800. km t₂ = --------------------- = 1.80 h 1000. km h^-1 t = t₁ + t₂ = 3.00 h + 1.80 h = 4.80 h d = d₁ + d₂ = 2400. km + 1800. km = 4200. km distance 4200. km average speed = ------------------- = --------------- = 875 km/h elapsed time 4.80 h	Assume the acceleration is instantaneous. Distance and velocity do not require a conversion of units. pay attention to significant figures. Significant figures and Significant Figure Calculations Total time is based on the sum of time for the two segments. Average speed is based on total distance divided by total time.

4.	At highway speeds, an Explorer is capable of accelerating 2.5 m s^-2. How long does it take to accelerate from 100. km/h to 150. km/h?
	/\v = v_f - v_i = 150. km/h - 100. km/h = 50. km/h km 1000 m 1 h 50. ------- x ------------ x ------------ = 14 m/s h 1 km 3600 s /\v 14 m s^-1 /\v = at => t = ----- = ----------------- = 5.6 seconds a 2.5 m s^-2	The units must be converted. Since the time is short, convert km/h into m/s. This can be done after calculating /\v to cut the conversions in half. pay attention to significant figures. Significant figures and Significant Figure Calculations

5.	A race car has a braking distance of 50. m at 130. km/h. What is the acceleration in m/s² and what is the g force?
	Forward is positive and backward is negative. Displacement and initial velocity are both positive. 130. km 1000 m 1 h ------------ x -------------- x ------------ = 36.1 m/s 1 h 1 km 3600 s v² - v_o² - v_o² - (36.1 m s^-1)² x = x_o + --------------- => a = ----------- = -------------------- 2a 2 x 2 x 50.0 m 1 g a = - 13.0 m s^-2 x ---------------- = - 1.33 g 9.80 m s^-2	Pick a reference frame. 1 g = 9.8 m s^-2 Solve for acceleration. v = 0 when braking x_o = 0 m km/h should be converted into m/s to match the other units. pay attention to significant figures. Significant figures and Significant Figure Calculations

6.	A world-class sprinter can burst out of the blocks to essentially top speed of 11.5 m/s in the first 15.0 m of the race. What is the average acceleration and time necessary to reach this speed?
	Forward is positive and backward is negative. v² - v_o² v² (11.5 m s^-1)² x = x_o + --------------- => a = ------- = -------------------- 2a 2 x 2 x 15.0 m a = 4.41 m s^-2 /\v 11.5 m s^-1 /\v = at => t = ------ = ----------------- = 2.61 s a 4.41 m s^-2	Pick a reference frame. v_o = 0 x_o = 0 m Solve for acceleration first.

7.	In coming to a stop, a car leaves skid marks on the highway 110. m long. Assuming a deceleration of 4.00 m/s², what was the speed of the car before braking?
	Forward is positive and backward is negative. x = 110. m a = - 4.00 m s^-2 v² - v_o² x = x_o + --------------- or v² = v_o² + 2a(x - x_o) 2a v_o = ( - 2 ax )^1/2 = ( - ( - 4.00 m s^-2) x 110. m )^1/2 = 21.0 m/s	Pick a reference frame. v = 0 when braking x_o = 0 m pay attention to significant figures. Significant figures and Significant Figure Calculations

8.	A woman driving her car 60. km/h approaches an intersection just as the traffic light turns yellow. The yellow light last only 3.0 s before turning red and the car is 50. m from the beginning of the intersection. The intersection is 12 m wide and the car can decelerate 5.0 m/s² and can accelerate from 60. km/h to 90. km/h in 6.0 s. Ignoring the length of the car and reaction time, should the driver try to stop?
	Forward is positive and backward is negative. km 1000 m 1 h v_o = 60. -------- x -------------- x ----------- = 16.7 m/s h 1 km 3600 s km 1000 m 1 h v = 90. -------- x -------------- x ----------- = 25.0 m/s after a for 6.0 s h 1 km 3600 s v - v_o 0 - 16.7 m s^-1 time to stop = ---------- = ---------------------- = 3.34 s a - 5.0 m s^-2 x = distance to stop = v_ot + 1/2 at² x = 16.7 m/s x 3.34 s + 1/2 x ( - 5.0 m s^-2 ) x (3.34 s)² = 27.9 m The car can stop before the intersection. v - v_o 25.0 m s^-1 - 16.7 m s^-1 a to light = ---------- = ---------------------------------- = 1.38 m s^-2 t 6.0 s x = distance during yellow light = v_ot + 1/2 at² x = 16.7 m s^-1 x 3.0 s + 1/2 x 1.38 m s^-2 x (3.0 s)² = 56.3 m The car would have to travel 50. m + 12 m = 62 m to clear the intersection. It cannot clear the intersection in the allowed time. The car should stop to avoid a ticket.	Pick a reference frame. Convert all velocities into m/s. The car should try to stop if the car can stop before the intersection. Calculate time then distance to stop. The car can safely go through if the car is through the intersection before the light turns red. Calculate how far the car can travel during the yellow light. pay attention to significant figures. Significant figures and Significant Figure Calculations

9.	A stone is dropped from the top of a building. It is seen to hit the ground after 5.35 s. How high is the building? What was the velocity of the stone just before it hit?
	Down is positive and up is negative. x = 1/2 a t² = 1/2 x 9.80 m s^-2 x (5.35 s)² = 140. m v = at = 9.80 m s^-2 x 5.35 s = 52.4 m/s	Pick a reference frame. v_o = 0 m/s because the stone is dropped. pay attention to significant figures. Significant figures and Significant Figure Calculations

10.	A parachutist jumps from an airplane. A second parachutist jumps from the airplane 2.50 s later. How far apart are the parachutist when the second parachutist has reached 22.0 m/s?
	Down is positive and up is negative. /\v 22.0 m s^-1 t = ------ = ----------------- = 2.24 s for 2nd Parachutist a 9.80 m s^-2 t = 2.50 s + 2.24 s = 4.74 s for 1st Parachutist x_{first parachutist} = 1/2 a t² = 1/2 x 9.80 m s^-2 (4.74 s)² = 110. m x_{second parachutist} = 1/2 a t² = 1/2 x 9.80 m s^-2 (2.24 s)² = 24.6 m /\x = 110. m - 24.6 m = 85 m	Pick a reference frame. v_o = 0 m/s because they are jumping from the airplane. Calculate the time necessary for the second parachutist to reach a velocity of 12.0 m/s. Calculate how far the first parachutist has dropped in 2.50 s + time Calculatate how far the second parachutist drops in time. The difference is how far apart they are at this time. pay attention to significant figures. Significant figures and Significant Figure Calculations

11.	A baseball is thrown up vertically with a speed of 14.0 m/s from the edge of a 200.0 m high cliff. How much time passes before it hits the bottom? What is its speed just before hitting? What is the total distance traveled by the baseball.
	Down is positive and up is negative. v_o = - 14.0 m/s a = 9.80 m/s x = 200.0 m x = v_ot + 1/2 at² = ( -14.0 m/s) t + 1/2 x 9.80 m s^-2 t² = 200.0 m 200 = -14.0 t + 4.9 t² => 4.9 t² - 14.0 t - 200 = 0 Solve the Quadratic t = 7.98 s or t = - 5.12 s Since the ball cannot hit the bottom before it is tossed, the correct answer is 7.98 s. v = v_o + at = - 14.0 m s^-1 + 9.80 m s^{- 2} x 7.98 s = 64.2 m s^-1 v² - v_o² - ( - 14.0 m s^-1)² x = ------------ = -------------------------- = - 10.0 m (above cliff) 2a 2 x 9.80 m s^-2 Total distance traveled = 2 x 10.0 m + 200.0 m = 220.0 m	Pick a reference frame. You must calculate the maximum altitude to calculate the total distance traveled by the baseball. The maximum altitude is reached when v = 0 m/s. pay attention to significant figures. Significant figures and Significant Figure Calculations

12.	A person who is properly constrained by an over-the-shoulder seat belt has a good chance of surviving a car collision if the deceleration does not exceed 15 g’s. Assuming uniform deceleration, how much must the front end collapse if a crash brings the car to rest from 100. km/h?
	Forward is positive and backward is negative. 9.80 m s^-2 15 g x ----------------- = 150 m s^-2 1 g km 1000 m 1 h 100. ------ x ---------- x ----------- = 27.7 m s^-1 h 1 km 3600 s v² - v_o² - ( 27.7 m s^-1)² x = ------------ = -------------------------- = - 2.6 m 2a 2 x 150 m s^-2	Pick a reference frame. v = 0 when braking 1 g = 9.80 m s^-2 Convert v_o into m/s Assume the distance it takes the person to stop is the distance the front of the car must crumple. It cost much less to replace a car than it does to pay a major medical bill. Insurance companies have lower rates for cars with front ends that crumple and protect the passengers than for cars that do not have good crumple zones. pay attention to significant figures. Significant figures and Significant Figure Calculations

13.	A ball is thrown upward with an initial velocity of 65.0 m/s. What is the highest point reached by the ball? How long is the ball in the air?
	v = v_o + at = 0 - v_o - 65.0 m s^-1 t = ------- = ------------------- = 6.63 seconds a - 9.80 m s^-2 Y = v_{o vertical} t + 1/2 gt² Y = 65.0 m/s x 6.63 s + 1/2 x ( - 9.80 ms^-2) (6.63 s)² Y = 216 m Total time in the air = 2 x 6.63 seconds = 13.26 seconds	The highest point is reached when the velocity is zero. Up is positive and down is negative. It takes the same amount of time for the ball to return to the Earth as it did for the ball to reach its highest point.

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