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A particle starts from a point with a velocity of 6m/s and moves with an acceleration of -2m/s².show that the particle will be at the starting point after 6second?
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A particle starts from a point with a velocity of 6m/s and moves with ...
To determine whether a particle will return to the starting point after 6 seconds, we can use the equation of motion for uniformly accelerated motion:
x = x0 + v0t + (1/2)at^2
where x is the final position, x0 is the initial position (which is 0 in this case), v0 is the initial velocity (6 m/s), a is the acceleration (-2 m/s�), and t is the time (6 s).
Plugging in the given values, we get:
x = 0 + (6 m/s)(6 s) + (1/2)(-2 m/s�)(6 s)^2
x = 0 + 36 m + (-18 m)
x = -18 m
The particle will be 18 meter away from the starting point after 6 seconds.
This question is part of UPSC exam. View all Class 9 courses
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A particle starts from a point with a velocity of 6m/s and moves with ...


Proof that the particle will be at the starting point after 6 seconds:

Given data:

  • Initial velocity (u) = 6 m/s
  • Acceleration (a) = -2 m/s²
  • Time (t) = 6 seconds


Calculating the final velocity:

We know the formula for calculating final velocity (v) using initial velocity, acceleration, and time:

v = u + at

Substitute the given values into the formula:

v = 6 + (-2 * 6)

v = 6 - 12

v = -6 m/s

Calculating the displacement:

Next, we can calculate the displacement using the formula:

s = ut + (1/2)at²

Substitute the values of u, t, and a into the formula:

s = 6*6 + (1/2)(-2)(6)²

s = 36 - 36

s = 0 m

Conclusion:

Since the displacement is 0 meters, it means that the particle will be at the starting point after 6 seconds. The negative sign in the final velocity indicates that the particle has moved in the opposite direction of its initial motion and has come back to the starting point.


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