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Question: A particle starts moving in xy – plane at 2 m/s along x – axis. 2 second later, its velocity is 4 m/...

A particle starts moving in xy – plane at 2 m/s along x – axis. 2 second later, its velocity is 4 m/s in a direction making 600 with positive x – axis. Its average acceleration for this period of motion is:

A

5\sqrt { 5 } m/s2, along y – axis

B

3\sqrt { 3 } m/s2, along y – axis

C

5\sqrt { 5 } m/s2, at 600 with positive x – axis

D

3 m/s2, at 600 with positive x – axis

Answer

3\sqrt { 3 } m/s2, along y – axis

Explanation

Solution

a=VfViΔt\vec { a } = \frac { \vec { V } _ { f } - \vec { V } _ { i } } { \Delta t }

= [4cos60i^+4sin60j^][2i^]2\frac { [ 4 \cos 60 \hat { i } + 4 \sin 60 \hat { j } ] - [ 2 \hat { i } ] } { 2 } = (2i^+23j^)2i^2\frac { ( 2 \hat { \mathrm { i } } + 2 \sqrt { 3 } \hat { \mathrm { j } } ) - 2 \hat { \mathrm { i } } } { 2 } = 3j^\sqrt { 3 } \hat { j }

m/s2