Motion in one dimension?

Question

Velocity and acceleration of a particle at some instant of time are v =(3i + 4j) m/s and a = -(6i + 8j) m/s^2 respectively. At the same instant particle is at origin. Maximum x-coordinate of particle will be:

a) 1.5 m

b)0.75 m

c) 2.25 m

d) 4.0 m

i , j - unit vectors in x and y axes respectively.

Answer 1

You need the maximum x-coordinate.

So you need to consider only the components along i.

Now the velocity is in the +X direction and the acceleration is in the -X direction. From this we can deduce that the particle will move in the +x direction for some time, while it will continue to slow down. It will finally stop and then move in the -x direction with increasing speed.

So the maximum x coordinate is when the particle is at rest.

u = 3

a = -6

v = 0

Use the 3rd equation of motion :

s = (v^2 - u^2)/2a = 0.75 m

Hope this helps.

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