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One-Dimensional Kinematics Problems and Answers, Lecture notes of Kinematics

Kuwait University (KU)Kinematics

A series of problems and answers related to one-dimensional kinematics, including questions about particles moving in one dimension, kinematic equations, and graphs of position, velocity, and acceleration versus time. The problems involve calculating the velocity, acceleration, and distance traveled by objects, as well as identifying which particle is moving with constant or non-zero acceleration.

Typology: Lecture notes

2021/2022

Uploaded on 08/01/2022

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One-Dimensional Kinematics Horizontal Motion ANSWER KEY
1
1. ____B.____ A car travels 30 kilometers at an average speed of 60 kilometers per hour and
then 30 kilometers at an average speed of 30 kilometers per hour. The average speed of the car
over the 60 kilometers is
(A) 35 km/h (B) 40 km/h (C) 45 km/h (D) 50 km/h (E) 53 km/h
Ans. The answer is not 45 km/hr as might seem initially. The problem is that both paths donโ€™t
take the same amount of time. Find the total time for each. V = ๐’”
๐’• ๏‚ฎt = ๐’”
๐’— = ๐Ÿ‘๐ŸŽ ๐’Œ๐’Ž
๐Ÿ”๐ŸŽ ๐’Œ๐’Ž/๐’‰๐’“ = 0.5
hrs. and t = ๐’”
๐’— = ๐Ÿ‘๐ŸŽ ๐’Œ๐’Ž
๐Ÿ‘๐ŸŽ ๐’Œ๐’Ž/๐’‰๐’“ = 1 hr. The total time is 1.5 hrs. Vavg = ๐’”
๐’• =๐Ÿ”๐ŸŽ ๐’Œ๐’Ž
๐Ÿ.๐Ÿ“ ๐’‰๐’“๐’”.= 40 km/hr.
Questions 2-4 relate to five particles that start at x = 0 at t = 0 and move in one dimension independently
of one another. Graphs of the velocity of each particle versus time are shown below.
2. ____E.____ Which particle is farthest from the origin at 2 s?
(A) A (B) B (C) C (D) D (E) E
Ans. Particle E had the largest velocity for the longest
time moving the entire time from the origin at ๏€ญ1 m/s.
3. ____A.____ Which particle moves with a constant nonzero
acceleration?
(A) A (B) B (C) C (D) D (E) E
Ans. The acceleration is the slope of the velocity graph.
A is the only graph that shows a line with a constant
non-zero slope.
4. ____C.____ Which particle is in its initial position at t = 2 s?
(A) A (B) B (C) C (D) D (E) E
Ans. Particle C has travelled at +1 m/s for one second
and ๏€ญ1 m/s for one second. Therefore, he is back where
he started.
5. ____C.____ The graph above shows the velocity versus time for an object moving in a straight
line. At what time after time = 0 does the object again pass through its initial position?
(A) Between 0 and 1 s (D) 2s
(B) 1 s (E) Between 2 and 3 s
(C) Between 1 and 2 s
Ans. Since we are given the velocity vs. time graph, the area under the curve will give us the distance
travelled vs. time. When the negative area under the curve between t = 0 s and t = 1 s is the same as
the positive area under the curve from t = 1 to our answer, then the object will have passed once
again through the origin.
The distance travelled from t =0 s to t = 1 s is A = ยฝbh = ยฝ(1)(๏€ญ1) = ๏€ญยฝ m. By t = 2 seconds, the
area has already become A = ยฝbh = ยฝ(1)(2) = 1 m so the car is already moving away from initial
position in a positive direction.
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1. ____ B. ____ A car travels 30 kilometers at an average speed of 60 kilometers per hour and

then 30 kilometers at an average speed of 30 kilometers per hour. The average speed of the car over the 60 kilometers is (A) 35 km/h (B) 40 km/h (C) 45 km/h (D) 50 km/h (E) 53 km/h

Ans. The answer is not 45 km/hr as might seem initially. The problem is that both paths donโ€™t

take the same amount of time. Find the total time for each. V =

๐’”

๐’• ๏‚ฎ t =^

๐’”

๐’— =^

๐Ÿ‘๐ŸŽ ๐’Œ๐’Ž

hrs. and t =

๐’”

๐’— =^

๐Ÿ‘๐ŸŽ ๐’Œ๐’Ž

๐Ÿ‘๐ŸŽ ๐’Œ๐’Ž/๐’‰๐’“ = 1 hr. The total time is 1.5 hrs.^ Vavg^ =^

๐’”

๐’• =^

๐Ÿ”๐ŸŽ ๐’Œ๐’Ž

๐Ÿ.๐Ÿ“ ๐’‰๐’“๐’”. = 40 km/hr.

Questions 2-4 relate to five particles that start at x = 0 at t = 0 and move in one dimension independently of one another. Graphs of the velocity of each particle versus time are shown below.

  1. ____ E. ____ Which particle is farthest from the origin at 2 s? (A) A (B) B (C) C (D) D (E) E Ans. Particle E had the largest velocity for the longest time moving the entire time from the origin at ๏€ญ 1 m/s.
  2. ____ A. ____ Which particle moves with a constant nonzero acceleration? (A) A (B) B (C) C (D) D (E) E Ans. The acceleration is the slope of the velocity graph. A is the only graph that shows a line with a constant non-zero slope.
  3. ____ C. ____ Which particle is in its initial position at t = 2 s? (A) A (B) B (C) C (D) D (E) E Ans. Particle C has travelled at +1 m/s for one second and ๏€ญ 1 m/s for one second. Therefore, he is back where he started.

5. ____ C. ____ The graph above shows the velocity versus time for an object moving in a straight

line. At what time after time = 0 does the object again pass through its initial position? (A) Between 0 and 1 s (D) 2s (B) 1 s (E) Between 2 and 3 s (C) Between 1 and 2 s Ans. Since we are given the velocity vs. time graph, the area under the curve will give us the distance travelled vs. time. When the negative area under the curve between t = 0 s and t = 1 s is the same as the positive area under the curve from t = 1 to our answer, then the object will have passed once again through the origin. The distance travelled from t =0 s to t = 1 s is A = ยฝbh = ยฝ(1)( ๏€ญ 1) = ๏€ญ ยฝ m. By t = 2 seconds, the area has already become A = ยฝbh = ยฝ(1)(2) = 1 m so the car is already moving away from initial position in a positive direction.

Questions 6-7 relate to five objects that are moving in parallel straight-line paths. The objects all cross a starting line at the instant a clock is started. The distances from the starting line in meters after 1, 2, 3, 4, and 5 seconds are as follows:

6. ____ B. ___Which object is moving with

zero acceleration? (A) A (B) B (C) C (D) D (E) E Ans. The distance increases by the same amount in Object C. This means the object is cruising and not accelerating.

7. ____ D. ___Which object has constant nonzero acceleration and appears to have started from rest?

(A) A (B) B (C) C (D) D (E) E

Ans. Object Dโ€™s distances are increasing by a steady amount every second: First by 4 m, then 6 m, then 8 m. This suggests the velocity is changing at a constant rate, hence, a non-zero acceleration.

8. ____ B. ____ Which graph of position x versus time t best represents a moving object with positive

velocity and positive acceleration, both in the x direction?

Ans. The objectโ€™s position is increasing as an upward facing curve. This suggests the distances are getting bigger in every time unit and so the velocity is increasing which means the object has acceleration.

9. ____ D. ____ The graph at right shows acceleration a as a function of time t for

a particle moving in a straight line. Graphs of speed v versus time t that are consistent with the a versus t graph above include which of the following?

(A) I only (B) III only (C) I and II only (D) II and III only (E) I, II, and III

Ans. Both II and III show a non-zero increasing linear velocity. This coincides with a graph of a positive, constant acceleration as shown in the graph at right. Graph I shows a velocity that is constant and not increasing. The acceleration for this graph would be a horizontal line but at a height of zero.

Time (seconds) Object 1 2 3 4 5 (A) 1 m 1 m 2 m 2 m 3 m (B) 1 m 2 m 3 m 4 m 5 m (C) 1 m 4 m 9 m 16 m 25 m (D) 4 m 10 m 18 m 28 m 40 m (E) 6 m 11 m 15 m 18 m 20 m

13. Problem: Sketch the Kinematic Graphs for Constant (Uniform) Acceleration

Stationary particle Particle moving with constant velocity

x vs t v vs t a vs t x vs t v vs t a vs t

Particle moving with constant non-zero acceleration

x vs t v vs t a vs t

Problem: Kinematic Graphs (CM-1988)

(E)

14. (E)

Problem: Kinematic Graphs (CM-1998)

  1. ____ D. ____ The graph at left shows the velocity v as a function of time t for an object moving in a straight line. Which of the following graphs shows the corresponding displacement x as a function of time t for the same time interval?

Expl. Ans. D. The velocity is increasing in the first

segment so the object is accelerating. This corresponds to a curved displacement graph

opening upwards. In the second segment the velocity is staying the same. This corresponds to

a displacement graph that is increasing in a linear fashion. The third segment is decreasing

velocity , or negative acceleration. This corresponds to a curved graph but opening

downwards.

FREE RESPONSE PRACTICE PROBLEMS

DIRECTIONS: ANSWER EACH OF THE FOLLOWING PROBLEMS! BE SURE TO

SHOW ALL WORK CAREFULLY!

I. A student is running to catch the campus shuttle bus, which is stopped at the bus stop. The student is running at a constant velocity of 6 m/s; she cannot run any faster. When the student is still 80 m from the bus, it starts to pull away. The bus moves with a constant acceleration of 0.2 m/s^2. a) For how much time and how far will the student have to run before she overtakes the bus? Ans. The displacement of the student is given by xstudent = vot = 6t. The displacement of the bus is given by xbus = xo + ยฝat^2 = 80 + ยฝ(0.2 m/s^2 )t^2 = 80 + 0.1t^2. If we want to know when the student catches up with the bus, we should set the two displacement equations equal to each other and solve for t. xstudent = xbus 6t = 80 + 0.1t^2 0 = 0.1t^2 โ€“ 6t + 80 Solving for t using your graphing calculator: t = 20 s At t = 20 seconds the student has moved x = = vot = 6(20 s) = 120 m. b) When she reaches the bus, how fast will the bus be traveling? Ans. The velocity formula for the bus is vbus = aยทt = 0.2 m/s^2 (20 s) = 4 m/s c) Sketch a graph showing x(t) for both the student and the bus. Take x = 0 as the initial position of the student. (The same axis would be useful.)

t

x

(D)