1.

An object is moving along+ve x-axis with a uniform acceleration of 4 ms^(-2)`. At time ` t=0`. X= 4 m` and ` v=2 ms^(-1)``. (a) What will be the velocity and position of the object at time ` t=3 s? (b) What will be the position of the object when it has a velocity ` 8 ms^(-1)` ?

Answer» Here, ` x (0) = 4 m, v (0) = u = 2 ms^(-1)`,
` a = 4 ms^(-2)`
(a) When ` t= 3 s,`
` v= u + at = 2 + 4 xx 3 14 am^(-1)`
Ppsotion of object at time ` t=3 s`,
` x=x_0 + ut = 1/2 at^2 = 4 + 2 xx 3 + 1/2 xx 4 xx 3^2
= 28 m`
(b) When ` u= 8 ms^(-1)`, posotion ` X (t) of the object is giben by ltbRgt ` `v^2 - u^2 2 a ( x- x_0)`
or ` x=x_0 + ( v^2-u^2)/( 2a) = 4 + ( 8^2 - 2^2)/( 2 xx 4)`
` = 4 + 15 //2 = 11. 5 m`.


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