2.1 Motion study "b"
Acceleration
In tracing the motion of Mr Samer's car, we notice that the speed of car does not remain uniform throughout the journey . We say that the Car is either accelerating (i.e. getting faster) or decelerating (i.e. getting slower).
Acceleration is defined as the rate of change of velocity.
The SI unit of acceleration is meter per square second (m/s2).
Mathematically, it is represented by the formula:
Acceleration = change in velocity (m/s)
Time taken (s)
= Final velocity - Initial velocity
Time Taken
Uniform Acceleration
The acceleration of an object is uniform when the rate of change of its velocity remains Constant. For example, a uniform acceleration of 2 m/s2 means that the velocity of the object changes y 2 m/s every one Second later.
Pause & think
An Object moves in a circular path, but its speed is uniform. Does it have an acceleration?
Let's consider the motion of a car accelerating at 2 m/s2. If the Car starts from rest, its velocity = 0 m/s at time t=0 s. One Second later, its velocity increase to 2 m/s During the next one second, its velocity will have increased another 2 m/s to become 4 m/s. At the end of the third second, its velocity is 6 m/s. At each one second time interval, its velocity increases at the Same amount of 2 m/s every one second. We say that the car accelerates uniformly. It has uniform acceleration.
Worked Example 1.2
A car accelerates uniformly from rest to a velocity of 15 m/s in 3 s.
(a) Calculate the acceleration of the car
(b) Predict the value of the velocity of the car at the end of 5 s if its acceleration remains constant. Explain your prediction.
Solution
(a) As the car starts from rest, its initial velocity = 0 m/s.
acceleration = Final velocity - initial velocity
time taken
= 15 - 0 = 5 m/s2
3
(b) The acceleration of the car is 5 m/s2, it means that every 1 s later its velocity increases by 5 m/s. Hence, the velocity of the car at the end of 5 s is ( 15 + 5+5) = 25 m/s.
Alternatively, We Know from (a), its initial velocity is 0 m/s and its acceleration is 5 m/s2; at the end of 5 s, its velocity can be calculated using the formula
acceleration = velocity / time taken
velocity = 5 × 5 = 25 m/s
Test yourself "on Facebook"
A skate boarder slides down a slope from rest with a uniform acceleration of 3 m/s2. What is his speed at the end of 4 s?
Speed-Time Graph
On another occasion, Bara observes the speedometer in his father's car at 1-minute intervals and records the speed "v" against time "t" in the table below. This data traces the journey from his home to the traffic junction just before the car hits the highway.
We Can easily translate the information above into a graph of speed "v" against time "t" to show the motion of the Car.
During the first 2 minutes, i.e. from t = 0 min. to t = 2 min, the car accelerates uniformly. The speed of the car increases at a constant rate of 40 km/h every 1 minute. Then the speed of the car becomes uniform at 80 km/h during the next 3 minutes, from t = 2 min. to t = 5 min. During this period, the speed of the car doesn't change. The acceleration of the car is zero.
At t = 5 min, the car accelerates again for the next 2 minutes. However this time the car accelerates non-uniformly as the increase in speed is not constant. At the end of t =7 min, the Car slows down or decelerates and finally comes to rest after 8 minutes into its journey.
= Final speed - Initial Speed = Acceleration
Time interval
Area under Speed- Time Graph
suppose Bara' father passes the traffic junction, and is able to accelerate uniformly to a speed of 36 km/h, or 10 m/s in 4 sm and maintains this uniform speed of 10 m/s for the next 4 s before bringing the car to rest uniformly.
**Note: Unit conversion 36 km/h = 3600 m = 1 m/s
3600 s
We can represent this part of the journey by the speed - time graph below
To obtain the total distance traveled by the car during this journey, we calculate the area under the speed -time graph;
Total distance traveled = Area "A" + Area "B" + Area "C" = Area of Trapezium
= 0.5 * (10+4) * 10
= 70 m
Acceleration of Speed - Time Motion
To obtain the acceleration "a" of the car during the first four seconds and the last two seconds, we calculate the slopes under the graph.
a (t = 0 to t = 4) = slope of positive slope = 10 ÷ 4 = 2.5 m/s2
a (t = 8 to t = 10) = slope of negative slope = -10 ÷ 2 = -5 m/s2
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