How to Find Acceleration Using a Velocity Time Graph

What Is Acceleration

 

  • Rate of change of velocity is called acceleration. It is a vector quantity
    \(i.e.\text{    }a=\frac{v-u}{t}\)
    where u is the initial velocity of the object, v is its final velocity and t is the time taken.
  • Unit of acceleration = m/s2 or ms-2
  • If the velocity of a body decreases, then it will experience a negative acceleration which is called deceleration or retardation.
  • Figure shows a car moving along a straight line. The speedometer of the car shows that it is moving with increasing velocity. The car is accelerating.
    Analysing Linear Motion 3
  • We say that an object is undergoing a deceleration or retardation when it is slowing down. The rate of change of velocity of the object then has a negative value. Figure shows a car decelerating. The speedometer of the car shows that it is moving with decreasing velocity.

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Types of acceleration

Uniform acceleration: If a body travels in a straight line and its velocity increases by equal amounts in equal intervals of time then it is said to be in state of uniform acceleration.
e.g. motion of a freely falling body.

Non uniform acceleration: A body has a non-uniform acceleration if its velocity increases by unequal amounts in equal intervals of time.

Instantaneous acceleration: The acceleration of a body at any instant is called its instantaneous acceleration.

Acceleration is determined by the slope of time-velocity graph.
\(\tan \theta =\frac{dv}{dt}\)

  1. If the time velocity graph is a straight line, acceleration remains constant.
  2. If the slope of the straight line is positive, positive acceleration occurs.
  3. If the slope of the straight line is negative, negative acceleration or retardation occurs.

Analysing MotionAnalysing Motion 1

  1. Figure shows a set-up of apparatus to analyse motion in the laboratory.
  2. (a) A ticker timer is an apparatus that gives a permanent record of motion for further analysis. When connected to an alternating current (a.c.) power supply (usually 12 V), it vibrates at a frequency of 50 Hz.
    (b) The ticker timer makes a series of dots at a rate of 50 dots per second on a piece of ticker tape as it is pulled through the timer by a trolley. Therefore, the time interval of a dot and the next dot which is also known as one tick is 1/50 or 0.02 s.
    (c) The distance between two dots is equal to the distance travelled by the trolley during the time interval between the dots.
    (d) The ticker tape can be analysed to determine the time, displacement, average velocity, acceleration and type of motion of an object.
  3. The ticker tape can be cut into strips of equal time (equal number of ticks) and pasted together to form a chart for analysing the motion of a trolley.
  4. Figure shows three charts formed from strips of ticker tape, each consisting of ten ticks.Analysing Motion 2
  5. For motion with uniform acceleration or deceleration, its value can be determined by analysing the chart. Figure shows a chart formed from strips of ticker tape with ten ticks each.
    Analysing Motion 3
    Analysing Motion 4
    Analysing Motion 5

Acceleration Using A Velocity Time Graph Example Problems With Solutions

Example 1. A van accelerates uniformly from a velocity of 10 m s-1 to 20 m s-1 in 2.5 s. What is the acceleration of the van?
Solution: Initial velocity, u = 10 ms-1
Final velocity, v = 20 ms-1
Time taken, t = 2.5 s
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Example 2. A car travelling at 24 m s-1 slowed down when the traffic light turned red. After undergoing uniform deceleration for 4 s, it stopped in front of the traffic light. Calculate the acceleration of the car.
Solution:
 Initial velocity, u = 24ms-1
Final velocity, v = 0 ms-1
Time taken, t = 4 s
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Example 3. Time-velocity graph of a body is shown in the figure. Find its acceleration in m/s2.
Solution:    As it is clear from the figure,
At t = 0 s, v = 20 m/s
At t = 4 s, v = 80 m/s
Acceleration-Example-Problem
\(therefore \text{Acceleration,}a=\frac{\text{Change}\,\text{in}\,\text{velocity}}{\text{Timeint}\,\text{erval}} \)
\( =\frac{\Delta v}{\Delta t}=\frac{{{v}_{2}}-{{v}_{1}}}{{{t}_{2}}-{{t}_{1}}} \)
\( =\frac{(80-20)\,}{(4-0)}=15\text{ m/}{{\text{s}}^{\text{2}}} \)

Example 4. Time-velocity graph of a particle is shown in figure. Find its instantaneous acceleration at following intervals
Acceleration-Example-Problem-1
(i) at t = 3s
(ii) at t = 6s
(iii) at t = 9s
Solution:    (i) Instantaneous acceleration at t = 3s, is given by
a = slope of line AB = zero
(ii) Instantaneous acceleration at t = 6 s, is given by a = slope of line BC
\( =\frac{CM}{BM}=\frac{100-60}{8-4}=\text{ }10\text{ m/}{{\text{s}}^{\text{2}}} \)
(iii) Instantaneous acceleration at t = 9 s, is given by a = slope of line CD
\( =\frac{0-100}{10-8}=-50\text{ m/}{{\text{s}}^{\text{2}}} \)

Example 5. Starting from rest, Deepak paddles his bicycle to attain a velocity of 6 m/s in 30 seconds then he applies brakes so that the velocity of the bicycle comes down to 4 m/s in the next 5 seconds. Calculate the acceleration of the bicycle in both the cases.
Solution:    (i) Initial velocity, u = 0, final velocity,
v = 6 m/s, time, t = 30 s
Using the equation v = u + at, we have
\( a=\frac{v-u}{t} \)
substituting the given values of u, v and t in the above equation, we get
\( a=\frac{6-0}{30}=0.2\text{ m/}{{\text{s}}^{\text{2}}}\text{; }\!\!~\!\!\text{ } \)
which is positive acceleration.
(ii) Initial velocity, u = 6 m/s, final velocity,
v = 4 m/s, time, t = 5 s, then
\( a=\frac{v-u}{t}=\frac{4-6}{5}=-0.4\text{ m/}{{\text{s}}^{\text{2}}}\text{; }\!\!~\!\!\text{ } \)
which is retardation.
Note: The acceleration of the case (i) is positive and is negative in the case (ii).

Example 6. A trolley pulled a ticker tape through a ticker timer while moving down an inclined plane. Figure 2.10 shows the ticker tape produced.
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Determine the average velocity of the trolley.
Solution:
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Example 7. Figure shows ticker tapes produced from the motion of a trolley.
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Describe the type of motion of the trolley for each ticker tape.
Solution:
(a) The distances between two neighbouring dots are the same throughout the tape. Therefore, the trolley moved with uniform velocity.
(b) The distances between two neighbouring dots are increasing. Therefore, the trolley moved with increasing velocity The trolley was accelerating.
(c) The distances between two neighbouring dots are decreasing. Therefore, the trolley moved with decreasing velocity. The trolley was decelerating.
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Example 8. Figure shows a chart representing the movement of a trolley with uniform acceleration.
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Determine its acceleration.
Solution:
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Example 9. A trolley travelled down an inclined plane pulling along a ticker tape. Figure shows a chart formed by cutting and arranging the ticker tape into strips of ten ticks each.
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Determine the acceleration of the trolley.
Solution:
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Example 10. Figure shows a strip of ticker tape depicting the motion of a toy car with uniform acceleration.
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Determine the acceleration of the toy car.
Solution:
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