Can the average velocity of a body be the same as the instantaneous velocity? Explain.?

Answer 1

James forgot that when you accelerate or decelerate the instantaneous and average velocity can be the same in each case, in fact at some point they have to be equal.

The easy answer is that when the instantaneous velocity equals the average velocity; but this the idea situation under a uniform velocity with NO acceleration; beyond the original impulse.

What is going on is you need calculus to figure the velocity for the entire acceleration curve. You can use the standard algebra equation to figure out the velocity at one point on the curve; the instantaneous velocity.

The average velocity is the average value of the curve; where the curve hung out the most. So at some point the instantaneous velocity HAS to equal the average velocity. The graph can go above and below the average, but it has to cross the average velocity at some point.

Answer 2

Yes.

If the time the average is taken is infinitely small it is instantaneous. Also consider the fact that the instantaneous velocity fluctuates constantly and it fluctuates above and below the mean velocity. Therefore at some point the instantaneous equals the mean as it crosses from greater to less than.

Answer 3

Yes,it can but only If the body moves with constant velocity(acceleration=0) than initial velocity equal to final velocity equal to instantaneous velocity.

Answer 4

The projection of a vector A onto a vector B potential which you somewhat choose to be sure the vector ingredient of A interior the the direction of B. In different words, how most of the vector A lies interior the same direction because of the fact the direction that vector B is orientated or pointing to or heading. Mathematically: proj(A into B) = [(AoB)/|B²|]B the place, AoB => scalar dot made of A and B: AoB = |A||B|cos? and the perspective between the two given vectors: ? = fifty 3 deg. If A = i9.4 + j3.4 and B = i10.a million - j6.5 |A| = ?9.4²+3.4² = 10 |B| = ?10.a million²+6.5² = 12 cos53 = 0.6 AoB = 10*12*0.6 = seventy two proj(A into B) = [(AoB)/|B²|]B = (seventy two/a hundred and forty four)(i10.a million-j6.5) proj(A into B) = i5.05 - j3.25 b) proj(B right into a) = [(BoA)/|A|²]B comparable reasoning applies, ..try it. 2. If acceleration is uniform, then, definite the same reasoning will carry on with for the velocities: = (v1+v2)/2 = (v2 - v1)/(t2 - t1) good success!

Answer 5

yes..... but only if the velocity is constant. ie acceleration=0 the only way I can explain it is calculus..... i foget the highschool way. I will likely confuse you so i wont go there