Two particles are moving in a plane; one has the velocity component vix=1m/s, viy=5m/s and the other has the velocity components v2x=4m/s and v2y=3m/s if both started from the same point what is the angle between their paths

calculate the sines and cosines of the angles ....

sintheta1=5/sqrt26
cosTheta1=1/sqrt26

sinTheta2=3/5
cosTheta2=4/5 check all those...

Now,
Sin(A-B)=sianAcosB - CosBsinA

then compute aine (A B), and finally, the arcsin(A-B) which is the angle

V1 = 1 + 5i = 5.1m/s[78.7o].

V2 = 4 + 3i = 5m/s[36.9o].

78.7 - 36.9 = 41.8o between V1 and V2.

Both solutions of the impersonator of MathMate and Henry give the same correct answer of 41.820°.

By the way, for your information, impersonation is a serious offence at Jiskha, liable to be banned.

To find the angle between the paths of the two particles, we can use the concept of the dot product between two vectors. The dot product of two vectors is given by the formula:

A · B = |A| * |B| * cos(theta),

where A and B are the vectors, |A| and |B| are the magnitudes of the vectors, and theta is the angle between the vectors.

In this case, we can represent the velocities of the two particles as vectors:

Particle 1: v1 = (vix, viy) = (1 m/s, 5 m/s)
Particle 2: v2 = (v2x, v2y) = (4 m/s, 3 m/s)

Now, let's find the dot product of these two vectors:

v1 · v2 = (vix * v2x) + (viy * v2y)
= (1 m/s * 4 m/s) + (5 m/s * 3 m/s)
= 4 m^2/s^2 + 15 m^2/s^2
= 19 m^2/s^2

Next, let's find the magnitudes of the two vectors:

|v1| = √(vix^2 + viy^2)
= √(1^2 + 5^2)
= √(1 + 25)
= √26 m/s

|v2| = √(v2x^2 + v2y^2)
= √(4^2 + 3^2)
= √(16 + 9)
= √25 m/s
= 5 m/s

Now, let's substitute the values back into the dot product formula:

19 m^2/s^2 = √26 m/s * 5 m/s * cos(theta)

To find the value of cos(theta), we can rearrange the equation as follows:

cos(theta) = (19 m^2/s^2) / (√26 m/s * 5 m/s)
= 19 / (√26 * 5)
≈ 0.874

Finally, we can find the angle theta by taking the inverse cosine (cos^(-1)) of the value we found:

theta ≈ cos^(-1)(0.874)
≈ 29.7 degrees

Therefore, the angle between the paths of the two particles is approximately 29.7 degrees.