Velocity-Time Graph For Uniform Accelerated Motion


 
 
Concept Explanation
 

Velocity-Time Graph For Uniform Accelerated Motion

(ii) Velocity-Time Graph for Uniform Accelerated Motion

In uniform accelerated motion, the velocity changes with equal amount in equal interval of time. In this case the velocity -time graph is a straight line passing through the origin.

Interpretation:

The nature of the graph shows that velocity changes by equal amounts in equal intervals of time.

In equal interval of time 10 s, the change in velocity is 18 m/s, which remains same, that means the acceleration of the body is constant. Thus, for all uniformly accelerated motion, the velocity -time graph is a straight line.

Calculation of distance or magnitude of displacement :To determine the distance moved by the car from its velocity-time graph.

The area under the velocity-time graph gives the distance (magnitude of displacement) moved by the car in a given interval of time. Therefore, S = area of ABCDE

           = Area of the rectangle ABCD + Area of triangle ADE

          large =ABtimes BC +frac{1}{2}(ADtimes DE)

Example: The velocity -time graph of an ascending passenger lift is shown iin figure below. What is the acceleration of the lift

(i) during the first two seconds?

(ii) between 2nd and 10th second?

(iii) during the last two seconds?

SOLUTION  (i) Case I

large Delta v=4-0=4m/s,

large Delta t=2-0=2s,

large a_{1}=?

large therefore a_{1}=frac{Delta v}{Delta t}=frac{4}{2}=2;;;m/s^{2}

(ii) Case II

large Delta v=4.6-4=0.6m/s, Delta t=10-2=8s,a_{2}=?

large because a_{2}=frac{Delta v}{Delta t}

           large =frac{0.6}{8}=0.075;m/s^{2}

(iii) Case III

large Delta v=0-4.6=-4.6m/s,Delta t=12-10=2s,a_{3}=?

large therefore a_{3}=frac{Delta v}{Delta t}=frac{-4.6}{2}=-2.3;m/s^{2}

Negative sign shown retardation.

 

 
 
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