The position of a body in one dimensional motion varies with time as x=24t -3t² . What is the nature of its motion? Find a) its velocity after 4 sec. b) Its acceleration and c) its average velocity from t=0 during the interval when the velocity is positive.

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Solution

Given,
$x = 24 t - 3 t^{2} v = \frac{dx}{dt} = 24 - 6 t a = \frac{dv}{dt} = - 6 m / s^{2} The body is moving with uniform acceleration . (a) Velocity after 4 \sec v = 24 - 6 \times 4 = 0 m / s (b) Acceleration = - 6 m / s^{2} (c) Since velocity after t = 1 \sec v = 24 - 6 \times 1 = 18 m / s (positive value) At t = 0 Position, x = 24 \times 0 - 3 \times 0 = 0 At t = 1 x' = 24 \times 1 - 3 \times 1^{2} = 21 m Displacement in this time interval, x' - x = 21 - 0 = 21 m Average velocity = \frac{Displacement}{Time interval} = \frac{21}{(1 - 0)} = 21 m / s$

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The position of a body in one dimensional motion varies with time as x=24t -3t2 . What is the nature of its motion? Find a) its velocity after 4 sec. b) Its acceleration and c) its average velocity from t=0 during the interval when the velocity is positive.

The position of a body in one dimensional motion varies with time as x=24t -3t² . What is the nature of its motion? Find a) its velocity after 4 sec. b) Its acceleration and c) its average velocity from t=0 during the interval when the velocity is positive.