Question

(8c4p 17) A particle leaves the origin with an initial velocity v 3.28, in m/s. Įt experiences a constant acceleration a =-1.00i-0.90j, in m/s. What is the velocity of the particle when it reaches its maximum x coordinate? i-component of velocity? 0.00 m/s You are correct. Your receipt no. is 166-4230 revious Tries j-component of velocity? -2.95 m/s Incorrect. Tries 5/5 Previous Tries When does it reach its maximum x coordinate? -2.952 ms Submit Answer You have entered that answer before Incompatible units. No conversion found between "m/s" and the required units. Tries 1/5 Previous Tries Where is the particle at this time? i-component of position? 2.952 m/s 2 Submit Answer Incompatible units. No conversion found between "n/s 2" and the required units. Tries 0/5 Previous Tries j-component of position? -9.486j 2 Submit Answer Unable to interpret units. Computer reads units as "j2". Tries 0/5 Previous Tries

Answer 1

First we want to find the time in which the particle changes from a positive X speed, to a negative x speed, so we're looking for where x = 0.

If we use V = V0+at using initial velocity and acceleration, we get:

0 = 3.28 + -1(t)

t = 3.28

So, it changes direction along the x axis at 0, and that's its max x. To find velocity, we use time and it's acceleration along the y axis, so...

V = V0+at

V = 0 + (0.9)(3.28)

V = 2.952 (Answer)

When does it reach the max x-coordinate?

t = 3.28 s (answer)

We need to use the position equation twice, one for x and y.

X = X0+V0t+1/2at^2

X = 0+3.28(3.28)+1/2(-1)(3.28)^2

x = 5.3792m (answer)

Y = Y0+V0t+1/2at^2

Y = 0+0(3.28)+1/2(-0.9)(3.28)^2

Y = -4.84m (answer)

so, answer (b) is -4.5i+2.25j m