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How far does the runner whose velocity-time graph is shown in Fig. 2-34 travel in 16 s?

How far does the runner whose velocity-time graph is shown in Fig. 2-34 travel in 16 s?

Grade:11

4 Answers

Aditi Chauhan
askIITians Faculty 396 Points
9 years ago

Given below is the velocity-time graph for a runner who accelerates initially for 2 s, then moves with uniform speed for 8 s , then decelerating for another 2 s and finally moving with uniform speed for 4 s.

236-792_Capture3.JPG
It is important to note that the length of each box in the figure above is 2 seconds along the horizontal direction and 2 m/s along the vertical.
The displacement of the runner is given by the area under the velocity-time curve but the area under the magnitude of velocity and time graph gives the distance. This can be seen from the definition of distance in integral form given as:
s = \int vdt
To make it easy, divide the area into squares, rectangles and triangles (as shown in the figure above).
From the figure above, we have that the area under the curve is the sum of area of triangle ABC, rectangle BDEC, triangle DOF, rectangle OFGE and square FHIG respectively.
Now area of triangle ABC (say A1) is:

236-1243_Capture6.JPG
Area of rectangle BDEC (say A2) is:
236-1769_Capture7.JPG
Area of triangle DOF (say A3)is:
236-131_Capture4m.JPG
Area of rectangle OFGE (say A4)is:
236-2256_8mCapture.JPG
Area of square FHIG (say A5)is:
236-1408_11Capture.JPG
The total area (say s) under the curve is:

236-2345_100Capture.JPG
Therefore the runner has gone 100 m in 16 seconds.
Kushagra Madhukar
askIITians Faculty 628 Points
3 years ago
Dear student,
Please find the solution to your problem below.
 
The total distance covered can be obtained by calculating area under the v – t graph between t = 0 to t = 16 s
Now, The whole figure can be divided into 3 figures
A1 → a triangle from origin to t = 1s
A2 → a trapezium from t =1 s to t = 12 s above v = 4 m/s
A3 → a rectangle from t = 1 s to t = 16 s below v = 4 m/s
 
A1 = ½ x 1 x 4 = 2 m
A2 = ½ x ([12 – 1] + [10 – 2]) x (8 – 4)
     = ½ x 19 x 4
     = 38 m
A3 = (16 – 1) x 4
     = 60 m
 
Hence, The total distance covered = A1 + A2 + A3
                                                  = 2 + 38 + 60
                                                  = 100 m
 
Hope it helps.
Thanks and regards,
Kushagra
Kavita
13 Points
3 years ago
Just add the following areas:
Triangle from t(0) to t(2): 8 sq.m.
Square from t(2) to t(10): 64 sq.m.
Trapezoid from t(10) to t(12): 12 sq.m.
Square from t(12) to t(16): 16 sq.m.
Total= 100 sq.m.
Total = 100 SQ.m.
Gajendra Chilwal
13 Points
2 years ago
Distance traveled by the runner = 100 sq. metre 
Man I soo hungry .
He is aa fucking guy.
Runner masterbates daily .
And now he has become weak due to his bad habits.
He has ha gay in his house.

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