Question

an object moves in the +x direction at a speed of 48 m/s. As it passes through the origin, it starts to experience a constant acceleration of 4.0 m/s^2 in the -x direction.

a. how much time elapses before the object returns to the origin?

b. what is the magnitude of the velocity of the object when it returns to the origin?

c. what is the direction of the velocity of the object when it returns to the origin?

-x direction

+x direction

Answer 1

48 mis 4 m/s? rorigin) At point o, Initial velocity U= 48 m/s. Since acceleratim is acting against the direction og motion ( U ( S a = - 4 m/s2 Let at A the object stops it. =0. 2) V-48-hmis2 ... t = -4 m/s2 18 4 = 12 see. As the acceleration is still acting upon the body, it again staunts to move along the direction & motion re- in opposite direction of privious notions. Distance travelled obering 12 see by object- d=ut that? = 48X12 - 4x4 XL1212 : 288 m. considering motion from a too co 4m/s2 7 E o A x t

d=288 m, a = +4 m/s² Initial velocity lat A ) = 0 (0) d = utzt al d = othat? 2 288 = 1 x 4 xt² t2 = 12 sec. Hence total time tittz=24 sec. (b) Initial velocity uco Imotion along I - a=4 mlsz atoo) t = 12 sec V = utat = 4X12 = 48 m/s (1) Direction of Velocity when it returns to the origin will be same as that of acceleration's direction ie. along -x direction, = 12 sec