For part A, finding the initial velocity, I have seen people are taking the cos
ID: 1777153 • Letter: F
Question
For part A, finding the initial velocity, I have seen people are taking the cos and sin * 45 degrees? Please explain where the numbers are coming from for these formulas.
Thank you!
SI unit: meter per second per second, m/s FIND THE AVERAGE ACCELERATION ACTIVE EXAMPLE 3-2 A car is traveling northwest at 9.00 m/s. Eight seconds later it has rounded a corner and is now heading north at 15.0 m/s. What are the magnitude and direction of its average acceleration during those 8.00 seconds? Let the positive x direction be east, and the positive y direction be north. SOLUTION (Test your understanding by performing the calculations indicated in each step.) 1. Write out v 2. Write out v 3. Calculate : 4. Find aav 5. Determine av and 0: vi = (-6.36 m/s)2 + (6.36 m/s) Vf = (15.0 m/s) (6.36 m/s)2 + (8.64 m/s) av (0.795 m/s2)2 + (1.08 m/s2) aav = 1.34 m/s2, = 53.6" north of east YOUR TURN Find the magnitude and direction of the average acceleration if the same change in velocity occurs in 4.00 s rather than 8.00 s. (Answers to Your Turn problems are given in the back of the book.)Explanation / Answer
given
car travelling north west
let north be towards +y axis and east be towards +x axis
then speed of car, v = 9(-i + j)/sqroot(2) m/s
v = 6.363(-i + j) m/s
after 8 seconds
velocity u = 15j m/s
so average acceleration in t = 8 seconds be a
a = (u - v)/t = (15j + 6.363i - 6.363j)/8
a = 0.7954i + 1.079j m/s/s
so magnitude of a = |a| = sqroot(0.7954^2 + 1.079^2)
|a| = 1.3405 m/s/s
let angle thjis acceleration makles with east direction in counterclockwise direction be theta
then tan(tehta) = 1.079/0.7954
theta = 53.603 deg north of east
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