Question

A particle starts from the origin at t = 0 with an initial velocity of 5.0 m/s along the positive x axis. If the acceleration is (–3.0i + 4.5j) m/s^2, determine the velocity and position of the particle at the moment it reaches its maximum x coordinate.

$v = 5 + (5i - 3i t+4.5jt)\frac{m}{s}$

$t_{x-max}=\frac{-5}{-3} = 1.67s$
Can someone explain why when the particle reaches it maximum x coordinate?

Answer 1

Given that,

Initial velocity U =(5.0 m/s) i

The acceleration is a = (3i+4.5j)

The velocity is V = U + at = 5i +(3i+4.5j)t

The distance traveled is S = Ut +1/2 at $\wedge$ 2

S = (5t) i + (1.5i+2.25j)t $\wedge$ 2

then the x component is Sx = (5t)+1.5t $\wedge$ 2

when it reaches the maximum x distance dSx/dt = 0

hence 0 = 5+3t  

t =5/3

consider only magnitude by substituting the value of t in the velocity equation,

we get hence velocity is V = 10i +7.5 j

and position S = 125/6i +75/4 j