# 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 :

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 :

__Known :__

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.

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.

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

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…

Solution :

v_{t} = v_{o} + a t

v_{t }– v_{o} = a t

a = v_{t }– v_{o} / t

a = acceleration, v_{t }= final velocity, v_{o} = initial velocity, t = time interval.

Acceleration AB = (25 – 20) / (20 – 0) = 5 / 20 = 1/4 = 0.25 m/s^{2}

Acceleration BC = (45 – 25) / (40 – 20) = 20 / 20 = 1 m/s^{2}

Acceleration CD = (35 – 45) / (50 – 40) = 10 / 10 = 1 m/s^{2}

Acceleration DE = (25 – 35) / (70 – 50) = 10 / 20 = 1/2 = 0.5 m/s^{2}

Acceleration EF = (0 – 25) / (90 – 70) = 25 / 20 = 5/4 = 1.25 m/s^{2}

7. What is the distance traveled in 10 seconds

__Known :__

Initial velocity (v_{o}) = 0

Final velocity (v_{t}) = 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 :

v_{t} = v_{o} + a t

d = v_{o} t + ½ a t^{2}

v_{t}^{2} = v_{o}^{2} + 2 a d

*v _{t} = final velocity, v_{o} = initial velocity, a = acceleration, t = time interval, d = distance*

Acceleration (a) :

v_{t} = v_{o} + a t

20 = 0 + a (4)

20 = 4 a

a = 20 / 4

a = 5 m/s^{2}

Distance traveled in 10 seconds :

d = v_{o} t + ½ a t^{2}

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.

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/s^{2} 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.

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

A. 600 m

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.

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

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.

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

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 meters

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.

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 t_{1}, and between t_{1} and t_{2}

B. Between t_{1} and t_{2}, and between t_{2} and t_{3}

C. Between t_{2} and t_{3}, and between t_{1} and t_{2}

D. Between 0 and t_{1}, and between t_{2} and t_{3}

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 – t_{1 }= constant acceleration

t_{1} – t_{2} = constant velocity

t_{2} – t_{3} = constant acceleration

The correct answer is B.