Graphical of linear motion – problems and solutions

1. Graph distance (vertical) versus time (horizontal) as shown in the figure below. An object at rest shown by number…

Solution :

Graphical of linear motion – problems and solutions 1

Number 3, shown by a straight line.

2. Graph velocity (vertical) vs time interval (horizontal) of the motion at a constant acceleration. What is the magnitude of the acceleration according to the graph?

Solution :

Graphical of linear motion – problems and solutions 2

Graphical of linear motion – problems and solutions 3

Read :  Linear motion – problems and solutions

3. Someone traveled by car from town A to town B, shown by graph below. Vertical line as speed and horizontal line as time interval. What is the distance traveled by car during 30 minutes to 60 minutes.

Known :Graphical of linear motion – problems and solutions 4

Velocity (v) = 40 km/h

Time interval (t) = 60 – 30 = 30 minutes = 0.5 hours

Wanted : Distance

Solution :

Distance = speed x time interval

Distance = (40 km/hour)(0.5 hour)

Distance = 20 km

4. Graph velocity (vertical line) versus time interval (horizontal line) informs the motion of a car from rest then moves until stop for 8 seconds, as shown in figure below.

Graphical of linear motion – problems and solutions 5

What is the distance traveled by car from 5 seconds to 8 seconds?

Solution :

Area 1 = area of triangle CD = ½ (6-5)(40-20) = ½ (1)(20) = 10

Area 2 = area of triangle DE = ½ (8-6)(20-0) = ½ (2)(20) = 20

Area 3 = Area of square = (6-5)(20-0) = (1)(20) = 20

The distance traveled by car from 5 seconds to 8 seconds = 10 + 20 + 20 = 50 meters.

Read :  Magnetic force between two parallel wires – problems and solutions

5. The distance of the last 5 seconds according to graph below is ….

Graphical of linear motion – problems and solutions 6

Solution :

Area 1 = area of triangle = ½ (6-5)(40-20) = ½ (1)(20) = 10

Area 2 = area of rectangle = (9-5)(20-0) = (4)(20) = 80

Area 3 = area of triangle = ½ (10-9)(20-0) = ½ (1)(20) = 10

The distance of the last 5 seconds :

10 + 80 + 10 = 100 meters

6. Graph velocity vs time interval of the non-uniform linear motion of a car.

The same acceleration occurs at…Graphical of linear motion – problems and solutions 7

Solution :

vt = vo + a t

vt – vo = a t

a = vt – vo / t

a = acceleration, vt = final velocity, vo = initial velocity, t = time interval.

Acceleration AB = (25 – 20) / (20 – 0) = 5 / 20 = 1/4 = 0.25 m/s2

Acceleration BC = (45 – 25) / (40 – 20) = 20 / 20 = 1 m/s2

Acceleration CD = (35 – 45) / (50 – 40) = 10 / 10 = 1 m/s2

Acceleration DE = (25 – 35) / (70 – 50) = 10 / 20 = 1/2 = 0.5 m/s2

Acceleration EF = (0 – 25) / (90 – 70) = 25 / 20 = 5/4 = 1.25 m/s2

Read :  Nearsighted and farsighted – problems and solutions

7. What is the distance traveled in 10 seconds

Known :

Initial velocity (vo) = 0Graphical of linear motion – problems and solutions 8

Final velocity (vt) = 20 m/s

Time interval (t) = 4 seconds

Wanted :

The distance traveled in 10 seconds

Solution :

Three equations of the motion at a constant acceleration :

vt = vo + a t

d = vo t + ½ a t2

vt2 = vo2 + 2 a d

vt = final velocity, vo = initial velocity, a = acceleration, t = time interval, d = distance

Acceleration (a) :

vt = vo + a t

20 = 0 + a (4)

20 = 4 a

a = 20 / 4

a = 5 m/s2

Distance traveled in 10 seconds :

d = vo t + ½ a t2

d = (0)(10) + ½ (5)(10)2

d = ½ (5)(100)

d = (5)(50)

d = 250 meters

8. If an object is thrown vertically upward above the surface of earth, then which is the graph of acceleration experienced by the object.

Graph of linear motion problems and solutions 1

Solution

When an object moves vertically upward, the acceleration experienced by an object is acceleration due to gravity. The magnitude of acceleration due to gravity is 9.8 m/s2 and the direction of acceleration due to gravity is the to the center of Earth.

The constant acceleration is characterized by a straight line parallel to the axis t and perpendicular to the axis a.

The correct answer is D.

Read :  Bodies connected by the cord and pulley - application of Newton's law of motion problems and solutions

9. Based on the graph below, determine distance traveled by object for 20 seconds.

A. 600 mGraph of linear motion problems and solutions 2

B. 500 m

C. 200 m

D. 100 m

Solution

Distance traveled during 0 – 10 seconds = area of square + area of triangle

Area of square = (20-0)(10-0) = (20)(10) = 200 meters

Area of triangle = (1/2)(10-0)(40-20) = (1/2)(10)(20) = (5)(20) = 100 meters

Distance traveled during 0 – 10 seconds = 200 meters + 100 meters = 300 meters

Distance traveled during 10 – 20 seconds = area of triangle

Area of triangle = (1/2)(20-10)(40-0) = (1/2)(10)(40) = (5)(40) = 200 meters

Distance traveled during 0 – 20 seconds :

300 meters + 200 meters = 500 meters

The correct answer is B.

10. The motion of three objects, each illustrated by the graph below.

Graph of linear motion problems and solutions 3

Determine the correct statement for the motion of all three objects.

Graph of linear motion problems and solutions 4

Solution :

Graph of object 1 = Graph of acceleration (a) and time interval (t)

Based on the graph, object moves at a constant acceleration. Constant acceleration indicated by a straight line perpendicular to the axis of acceleration (a).

Graph of object 2 = Graph of velocity (v) and time interval (a)

Based on the graph, the object moves at a constant velocity. Constant velocity indicated by a straight line perpendicular to the axis of velocity (v).

Graph of object 3 = Graph of displacement (x) and time (a)

Based on the graph, displacement constant or object at rest.

The correct answer is A.

Read :  Boyle's law (constant temperature) - problems and solutions

11. The position of an object moving along the x-axis is shown by the following graph.

Graph of linear motion problems and solutions 5

The graph shows that the object moves at a constant velocity between the time interval….

A. 5-15 seconds and 25-35 seconds

B. 0-5 seconds and 35-40 seconds

C. 15-25 seconds

D. 0-5 seconds, 15-25 seconds and 35-40 seconds

Solution :

5-15 seconds and 25-35 seconds = Object’s displacement always constant or object at rest.

0-5 seconds, 15-25 seconds and 35-40 seconds = object moves at constant velocity.

The correct answer is D.

12. The following graph shows the speed of the time function of two objects moving straight from the same starting position. After moving for t seconds both objects have the same position change. Determine the time interval and displacement of the object.

A. 5 seconds and 50 metersGraph of linear motion problems and solutions 6

B. 5 seconds and 100 meters

C. 10 seconds and 50 meters

D. 10 seconds and 100 meters

Solution :

Graph 1 = constant acceleration

Time interval = 10 seconds

Distance = area of triangle = ½ (v)(t) = ½ (10-0)(20-0) = ½ (10)(20) = (5)(20) = 100 meters

Graph 2 = constant velocity

Time interval = 10 seconds

Distance = area of square = (v)(t) = (10-0)(10-0) = (10)(10) = 100 meters

The correct answer is D.

Read :  Normal force – problems and solutions

13. Based on graph below, the time interval when the object moves at constant acceleration and the time interval when the object is experiencing the greatest acceleration is ….

A. Between 0 and t1, and between t1 and t2Graph of linear motion problems and solutions 7

B. Between t1 and t2, and between t2 and t3

C. Between t2 and t3, and between t1 and t2

D. Between 0 and t1, and between t2 and t3

Solution :

If the graph is a straight line then the velocity is constant, if the graph is a slash then the acceleration is constant. The line is getting tilted, the acceleration is getting bigger.

0 – t1 = constant acceleration

t1 – t2 = constant velocity

t2 – t3 = constant acceleration

The correct answer is B.