Nonuniform linear motion – problems and solutions

Nonuniform linear motion – problems and solutions

1.

The table above shows data of three objects that travel the same distance at constant acceleration.

What are the final speed of object P and the initial speed of object Q?

Solution :

First, determine the distance traveled by object 3.

Distance traveled by object 3 :

Known :

Initial speed (v_o) = 0 m/s

Final speed (v_t) = 30 m/s

Acceleration (a) = 3 m/s²

Wanted : Distance

Solution :

v_t² = v_o² + 2 a s

v_t² – v_o² = 2 a s

30² – 0² = 2 (3) s

900 – 0 = 6 s

900 = 6 s

s = 900 / 6

s = 150 meters

The final velocity of object 1 :

Known :

Initial speed (v_o) = 20 m/s

Acceleration (a) = 4 m/s²

Distance (s) = 150 meters

Wanted : Final speed (v_t)

Solution :

v_t² = v_o² + 2 a s

v_t² = 20² + 2 (4)(150)

v_t² = 400 + 1200

v_t² = 1600

v_t = 40 m/s

The initial velocity of object 2 :

Known :

Final speed (v_t) = 50 m/s

Acceleration (a) = 3 m/s²

Distance (s) = 150 meters

Wanted : Initial speed (v_o)

Solution :

v_t² = v_o² + 2 a s

v_t² – 2 a s = v_o²

50² – 2(3)(150) = v_o²

2500 – 900 = v_o²

1600 = v_o²

v_o = 40 m/s

2. Three objects travel on a horizontal plane at constant acceleration. The three objects have the same acceleration. Data of the three object when travels in 10 seconds, shown in figure below.

Determine P and Q.

Solution :

First, determine the acceleration of object 1.

Acceleration of object 1 :

Known :

The initial speed (v_o) = 2 m/s

The final speed (v_t) = 22 m/s

Distance (s) = 120 meters

Wanted : Distance

Solution :

v_t² = v_o² + 2 a s

v_t² – v_o² = 2 a s

22² – 2² = 2 a (120)

484 – 4 = 240 a

480 = 240 a

a = 480/240

a = 2 m/s²

The initial speed of object 2 :

Known :

Acceleration (a) = 2 m/s²

Final speed (v_t) = 24 m/s

Distance (s) = 140 meters

Wanted : Initial speed (v_o)

Solution :

v_t² = v_o² + 2 a s

24² = v_o² + 2 (2)(140)

576 = v_o² + 560

576 – 560 = v_o²

16 = v_o²

v_o = 4 m/s

Distance of object 3 :

Known :

Initial speed (v_o) = 0 m/s

Final speed (v_t) = 20 m/s

Acceleration (a) = 2 m/s²

Wanted : Distance (s)

Solution :

v_t² = v_o² + 2 a s

20² = 0² + 2 (2) s

20² = 2 (2) s

400 = 4 s

s = 400/4

s = 100 meters

3. Determine the distance traveled by object in 40 seconds.

Solution :

Area 1 = area of rectangle = (20-0)(8-0) = (20)(8) = 160 meters

Area 2 = area of triangle = ½ (25-20)(8-0) = ½ (5)(8) = (5)(4) = 20 meters

Area 3 = area of triangle = ½ (30-25)(8-0) = ½ (5)(8) = (5)(4) = 20 meters

Area 4 = area of rectangle = (40-30)(8-0) = (10)(8) = 80 meters

The distance traveled in 40 seconds = 160 + 20 + 20 + 80 = 280 meters

4. The change of object’s speed in 2 seconds stated by graph below. Determine distance traveled by the object.

Solution :

Area 1 = area of triangle = ½ (5-0)(20-0) = ½ (5)(20) = (5)(10) = 50 meters

Area 2 = area of rectangle = (15-5)(20-0) = (10)(20) = 200 meters

Area 3 = area of triangle = ½ (20-15)(20-0) = ½ (5)(20) = (5)(10) = 50 meters

Distance traveled during 20 seconds = 50 + 200 + 50 = 300 meters

1. What distinguishes nonuniform linear motion from uniform linear motion?
   • Answer: Nonuniform linear motion involves a change in velocity over time, meaning there’s acceleration involved. In contrast, uniform linear motion implies that an object moves with a constant velocity and no acceleration.
2. How does the distance traveled by an object in nonuniform linear motion relate to the area under its velocity-time graph?
   • Answer: The distance traveled by an object in nonuniform linear motion is equal to the area under its velocity-time graph.
3. If an object’s acceleration-time graph is a straight horizontal line above the time axis, what does it indicate about the object’s motion?
   • Answer: It indicates that the object is undergoing constant positive acceleration. The object’s velocity is continually increasing at a steady rate.
4. Why can’t average velocity be simply calculated as the average of initial and final velocities in nonuniform motion?
   • Answer: For nonuniform motion, the velocity is not constant, so the actual displacement could be more or less than what is predicted by simply averaging the initial and final velocities. The correct method for nonuniform motion is to integrate the velocity over the given time interval or use kinematic equations that account for acceleration.
5. How would you describe the motion of an object whose velocity-time graph is a downward-sloping straight line?
   • Answer: A downward-sloping straight line on a velocity-time graph indicates that the object is moving with a constant negative acceleration, i.e., it’s decelerating or slowing down if it initially had a positive velocity.
6. In nonuniform motion, how does the instantaneous velocity at a particular moment relate to the slope of the displacement-time graph at that moment?
   • Answer: The instantaneous velocity at a particular moment in nonuniform motion is given by the slope or gradient of the displacement-time graph at that specific point.
7. What does a curve on a displacement-time graph suggest about the nature of an object’s motion?
   • Answer: A curve on a displacement-time graph indicates nonuniform motion, implying that the velocity of the object is changing (either increasing or decreasing) over time.
8. If an object’s displacement-time graph is parabolic and opens upward, what can you infer about its acceleration?
   • Answer: If the displacement-time graph is a parabola that opens upward, it suggests that the object is undergoing constant positive acceleration.
9. How does the acceleration of an object in nonuniform motion relate to the area under its velocity-time graph?
   • Answer: The change in velocity (which when multiplied by mass gives the change in momentum) of the object in nonuniform motion is equivalent to the area under its acceleration-time graph. It’s important to note that the velocity-time graph provides the change in velocity, not the acceleration directly.
10. What effect does negative acceleration (deceleration) have on the velocity of an object in nonuniform motion?

• Answer: Negative acceleration, often called deceleration, results in a decrease in the object’s velocity. If an object initially has a positive velocity and undergoes negative acceleration, its speed will decrease, and if deceleration continues, the object can change its direction of motion.