Final Speed of Particle: 2D Motion Problem

[blayman5]

Homework Statement

A particle moving at a velocity of 5.2 m/s in
the positive x direction is given an accelera-
tion of 5.3 m/s2 in the positive y direction for
9.9 s.
What is the final speed of the particle?
Answer in units of m/s

The Attempt at a Solution

I multiplied the acceleration by the time to get velocity.
I found the resultant velocity between the two vectors which was 52.7. Is that the same as the final speed?

 

[Melawrghk]

We're looking at a scenario where object's original speed in the y-direction is 0. Then they accelerated it in the y-direction, so the final speed will represent a line going diagonally through the first quadrant of a Cartesian coordinate system.
Acceleration is defined as change of speed over a period of time. So basically:
a = (v_f - v_i)/(change in t)
v_f = at + v_i = 5.3 m/s² * 9.9s + 0 m/s = 52.47m/s <= in y-direction.

Speed in x-direction is 5.2m/s. So its a triangle with one side being 52.47m/s and the other one 5.2m/s. Using regular Pythagorean theorem, the resultant speed (the hypotenuse, essentially) is 52.73m/s. According to significant digits, the right answer should be 53m/s.

So what you found is indeed the right answer. Good job!

 

[blayman5]

thanks Melawrghk

 

[blayman5]

A particlemoves in the xy plane with constant
acceleration. At time zero, the particle is at
x = 6 m, y = 1.5 m, and has velocity ~vo =
(3.5 m/s) ˆı + (−6 m/s) ˆ . The acceleration is
given by ~a = (5.5 m/s2) ˆı + (6 m/s2) ˆ.

What is the magnitude of the displacement
from the origin (x = 0 m, y = 0 m) after
2.5 s? Answer in units of m


I was wondering if you have a unit vector 17.25i+9j and a starting vector of 6.18466m of the displacement vector. how could you find the magnitude of the displacement after 2.5 sec.

I found the meters off of the resultant vector by multiplying the resultant by 2.5. Then i found the magnitude of the displacements. Where did i go wrong?