Motion in a Straight Line

CSEC Physics Syllabus - Effective for examinations from May - June 2015

Section A - Mechanics

Dynamics: Motion in a Straight Line

Specific Objective 4.1

define the terms: distance, displacement, speed, velocity, acceleration;

Specific Objective 4.2

apply displacement-time and velocity-time graphs;

Pixabay

Example 1

A boy rides his bicycle that records the distance traveled in meters. On a short ride he records the distance every 10 seconds and obtains the following readings: 

Time/s	0	10	20	30	40	60	70	80	90	100
Distance/m	0	20	40	60	80	200	260	280	280	280

a)    Plot a distance-time graph of his ride.

b)    What was his speed during the first 40 s?

c)    What was his speed between 40 s and 70 s from starting?

d)    What happened after 80 s?

    Solution

a) 

b)    His speed for the first 40 s is determined by the gradient of the graph during the specified time:

Gradient = Speed = Change in Distance ÷ Change in Time

Change in Distance = D₂ - D₁

Change in Time = t₂ - t₁

Using the following data points: 

D₂ = 80 m

D₁ = 0 

t₂ = 40 s

t₁ = 0

Speed = (80 m - 0) ÷ (40 s - 0)

Speed = 80 m ÷ 40 s

⸫ Speed = 2 m/s

c)    His speed between 40 s to 70 s is determined by the gradient of the graph during the specified time:

Gradient = Speed = Change in Distance ÷ Change in Time

Change in Distance = D₂ - D₁

Change in Time = t₂ - t₁

Using the following data points: 

D₂ = 260 m

D₁ = 80 m 

t₂ = 70 s

t₁ = 40 s

Speed = (260 m - 80 m) ÷ (70 s - 40 s)

Speed = 180 m ÷ 30 s

⸫ Speed = 6 m/s

d)    After 80 s, as seen in the distance-time graph, there was no increase in distance, which is a clear indication that he stopped his bicycle immediately. His speed went from an increasing value from 70 s to 80 s and then zero abruptly. 

Example 2

The graph below represents the velocity-time graph for a lift or elevator in a department store.

a)    Briefly describe the motion represented by OA, AB, and BC on the graph.

b)    Use the graph to calculate: 

        1.    the acceleration of the elevator

        2.    the deceleration of the elevator

        3.    the total distance traveled by the elevator.

    Solution

a)    The motion described by OA on the graph is as follows: 

At O, the elevator starts from rest, which means at time zero (0), its velocity is zero (0) and accelerates uniformly (uniform acceleration is identified by the straight line from O to A) in two (2) seconds to a velocity of 5 m/s to A. 

        The motion described by AB on the graph is as follows: 

From A to B, the elevator maintains its velocity of 5 m/s for one (1) second, which means the elevator did not experience any acceleration because there was no change in its velocity.  

        The motion described by BC on the graph is as follows: 

From B to C, the elevator decelerates uniformly (uniform deceleration is identified by the straight line from B to C) from its velocity from 5 m/s to 0 m/s in a time of four (4) seconds.  

b)

    1.    The acceleration of the elevator is experienced in the first two (2) seconds of its journey, from O to A:

Acceleration = Change in Velocity ÷ Change in Time

Change in Velocity = v₂ - v₁

Change in Time = t₂ - t₁

At O: 

v₁ = 0 m/s

t₁ = 0 s

At A: 

v₂ = 5 m/s

t₂ = 2 s

Acceleration = (5 m/s - 0 m/s) ÷ (2 s - 0 s)

Acceleration = 5 m/s ÷ 2 s

⸫ The Acceleration of the elevator = 2.5 m/s²

    

    2.    The deceleration of the elevator is experienced in the final four (4) seconds of its journey, from B to C:

Deceleration = - Acceleration = Change in Velocity ÷ Change in Time

Change in Velocity = v₂ - v₁

Change in Time = t₂ - t₁

At O: 

v₁ = 5 m/s

t₁ = 3 s

At A: 

v₂ = 0 m/s

t₂ = 7 s

Acceleration = (0 m/s - 5 m/s) ÷ (7 s - 3 s)

Acceleration = -5 m/s ÷ 4 s

⸫ Acceleration  = -1.25 m/s²

Recall Deceleration = - Acceleration

Deceleration = - (-1.25 m/s²)

⸫ The Deceleration of the elevator  = 1.25 m/s²

    

    3.    The total distance traveled by the elevator is determined by the area under the velocity-time graph. This could be solved by two methods:

Method 1: Divide the velocity-time graph into regular shapes (right-angle triangles and rectangles), calculate the area of each shape, and finally find the sum of all the areas. 

Area of the right-angle triangle formed by OA and the time axis: 

Area₁ = 1/2 × base × height

Area₁ = 1/2 × time from O to A × velocity at A

Area₁ = 1/2 × 2 s × 5 m/s

⸫ Area₁ = 5 m

Area of the rectangle formed by AB and the time axis: 

Area₂ = length × width

Area₂ = velocity at A or B × time from A to B 

Area₂ = 5 m/s × 1 s 

⸫ Area₂ = 5 m

Area of the right-angle triangle formed by BC and the time axis: 

Area₃ = 1/2 × base × height

Area₃ = 1/2 × time from B to C × velocity at B

Area₃ = 1/2 × 4 s × 5 m/s

⸫ Area₃ = 10 m

Therefore, 

The total distance traveled by the elevator = Area₁ + Area₂ + Area₃

The total distance traveled by the elevator = 5 m + 5 m + 10 m

⸫ The total distance traveled by the elevator = 20 m

Method 2: We use the area of a trapezium formula.

 Area = 1/2 × sum of the parallel sides × distance between them

Area = 1/2 × (time from OC + time from AB) × distance between OC and AB

Area = 1/2 × (1 s + 7 s) × 5 m/s

Area = 1/2 × 8 s × 5 m/s

⸫ Area = 20 m

Example 3

a)    What is meant by acceleration? 

An object was thrown vertically upwards and its height above the ground was measured at various times. The results obtained are shown in the table below. 

Time/s	0	1	2	3	4	5	6	7	8
Height/m	0	35	60	75		75	60	35	0

b)    Plot a graph of height on the y-axis against time on the x-axis. From the graph, find:

        1.    the maximum height reached

        2.    the time taken to reach this height.

c)    Using either or both of the answers from part b), calculate the initial velocity with which the object was thrown. Use g = 10 m/s²

    Solution

a)    Acceleration is defined as the change in velocity over a period of time. This change in velocity could be experienced as an increase or decrease in motion or a change of direction because velocity is a Vector Quantity.

b)

        1.    The maximum height reach was 83 m, as seen on the height against time graph above.

       2.    The time taken to reach this height was 4 s, as seen on the height against time graph above.

c)    When answering questions with Vector Quantities, we must assign which direction is positive or negative, because vector quantities depend on direction. Therefore, I assign every upward-moving vector as positive. This would cause gravity (g = 10 m/s²) to be negative because it is a downward-moving vector. 

Using the following equation:

Distance traveled = (initial velocity × time taken) + [1/2 × gravity × (time taken)²]

Where,

Distance traveled = s

initial velocity = u

time taken = t

gravity = g

s = ut + 1/2 × g × t²

83 m = u × (4 s) + 1/2 × (-10 m/s²) × (4 s)²

83 m = u × (4 s) + 1/2 × (-10 m/s²) × 16  s²

83 m = u × (4 s) + 1/2 × (- 160 m)

83 m = u × (4 s) - 80 m

 u × (4 s) = 83 m + 80 m

 u × (4 s) = 163 m

 u = 163 m ÷ 4 s

⸫ u = 40.75 m/s

Therefore, the initial velocity with which the object was thrown = 40.75 m/s