The relevant equation to use in this case is:
v = u + at ---> 1
where v is final velocity, u is initial velocity, a is acceleration, and t is time
Since u is 5m/s along the positive x axis so u = 5i. We are asked to find the time when maximum x coordinate is reached. This is also the time when the x component of the velocity reaches zero:
v=0
a (at maximum x) = -3.0i
Therefore
0 = 5i - 3i * t
t = 5/3 sec
Maximum x is reached at 5/3 sec
Equation 1 with x&y components is:
v = 5i + (-3i * t + 4.5j * t) ---> 2
Substituting t = 5/3 so get v:
v = 7.5 j m/s
Now we have the velocity, we need to find the position x. We know that:
v = dx / dt
Therefore equation 2 becomes:
dx / dt = 5i +(-3i * t + 4.5j * t)
By integrating we get:
x = (5t)i -((3.5t^2)/2)i +((4.5T^2)/2)j
Now substituting t = 5/3:
x =(125/35)i + (81/100)j
Answers:
v = 7.5 j m/s
x =(125/35)i + (81/100)j
The velocity of particle is 7.5j m/s and the position is 12.5 m towards the y-direction.
Given data:
The initial velocity of the particle is, u = 5.0 m/s.
The acceleration of particle is, [tex]a = (-3.0i + 4.5j) \;\rm m/s^{2}[/tex]
The given problem is based on the concept of velocity, which is expressed at the change in position with respect to time.
We have, acceleration at time instant t. Then for maximum x, the acceleration is expressed as,
a = (-3.0i + 0)
a = -3.0i
Now, obtain the time from the equation,
[tex]a = \dfrac{u}{t} \\\\t = \dfrac{5}{3} \;\rm s[/tex]
Now use the first kinematic equation of motion to obtain the velocity as,
v = ui + at
[tex]v = 5i +(-3i+4.5j) \times \dfrac{5}{3}\\\\v = 5i -5i+(4.5j \times \dfrac{5}{3})\\\\\v = 7.5j \;\rm m/s[/tex]
Now, position x is obtained as,
[tex]v = \dfrac{dx}{dt} \\\\\int dx =\int vdt\\\\x = v \times t\\\\x = 7.5j \times \dfrac{5}{3} \\\\x = 12.5j\;\rm m[/tex]
Thus, we can conclude that the velocity of particle is 7.5j m/s and the position is 12.5 m towards the y-direction.
Learn more about the velocity here:
https://brainly.com/question/17127206
