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Grade 12Mechanics

The velocity-time graph of a moving object is given in the figure. Find the maximum acceleration of the body and distance travelled by the body in the interval of time in which this acceleration exists.

Velocity-Time

Solution:

Acceleration is maximum when slope is maximum,

amax = (80-20)/(40-30) = 5m/s2

S = 20 m/s x 10 s + 1/2 x 6 m/s2 x 100s2 = 500m

My doubt is, in the above question, why can’t we find the distance travelled by the body by finding the area of v-t graph where the acceleration is maximum, i.e, distance=1/2 x 10 x 60=300.

Profile image of Kushi
11 Years agoGrade 12
Answers icon

1 Answer

Profile image of Paarth Arkadi
11 Years ago
Because distance travelled is not the area of the traingle that u mentioned in the graph.
But observe the bottom portion of the traingle which is a rectangle .
Hence if u want to solve the problem by taking the area under the graph in the interval then include area of the the triangle u had taken + area of the bottom partt ( the rectangle ) = 500m

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