If velocity v of a particle moving on a straight line as a function of...
Solution:
Given, velocity v of a particle moving on a straight line as a function of time t is given as v=5-t (m/s).
We know that, distance covered by a particle is given by the area under the velocity-time graph.
Step 1: Finding the distance covered by the particle in the first second.
- At t=0, v = 5 m/s
- At t=1, v = 4 m/s
- The area of the trapezium formed by v=5, v=4, t=0 and t=1 is:
- ((v1 + v2) / 2) * (t2 - t1)
- ((5 + 4) / 2) * (1 - 0)
- 4.5 m
Step 2: Finding the distance covered by the particle in the next second.
- At t=1, v = 4 m/s
- At t=2, v = 3 m/s
- The area of the trapezium formed by v=4, v=3, t=1 and t=2 is:
- ((v1 + v2) / 2) * (t2 - t1)
- ((4 + 3) / 2) * (2 - 1)
- 3.5 m
Step 3: Finding the distance covered by the particle in the next second.
- At t=2, v = 3 m/s
- At t=3, v = 2 m/s
- The area of the trapezium formed by v=3, v=2, t=2 and t=3 is:
- ((v1 + v2) / 2) * (t2 - t1)
- ((3 + 2) / 2) * (3 - 2)
- 2.5 m
Step 4: Finding the distance covered by the particle in the next second.
- At t=3, v = 2 m/s
- At t=4, v = 1 m/s
- The area of the trapezium formed by v=2, v=1, t=3 and t=4 is:
- ((v1 + v2) / 2) * (t2 - t1)
- ((2 + 1) / 2) * (4 - 3)
- 1.5 m
Step 5: Finding the distance covered by the particle in the next second.
- At t=4, v = 1 m/s
- At t=5, v = 0 m/s
- The area of the trapezium formed by v=1, v=0, t=4 and t=5 is:
-
If velocity v of a particle moving on a straight line as a function of...
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