Two velocities, fixed distance = time taken?

Question

Disclaimer: I am several decades out of school and I feel this is likely very simple. I just want verification.

Given a fixed distance $D$.

An object moves at speed $A$ $65\%$ of the time, and speed $B$ $35\%$ of the time.

What is the total time to cover distance $D$?

Is it simply

$$\frac D {A\cdot 0.65 + B \cdot0.35}$$ or is there more to it than that?

Answer 1

Yes, you are right:

Let $t$ be the (unknown) total time. Then

• the time of speed A is $\;0.65 t$, so the distance will be $A\cdot 0.65t$
• the time of speed B is $\;0.35 t$, so the distance will be $B\cdot 0.35t$

So the sum of these distances is $A\cdot 0.65t + B\cdot 0.35t = (A\cdot 0.65 + B\cdot 0.35)t$

From the other side, the total distance is $D$, so we obtain the equation

$$(A\cdot 0.65 + B\cdot 0.35)t = D$$

which you solved correctly.

Answer 2

In generalization, the velocity would be union of A and B velocities events,

P(v)=P(A) U P(B)

P(v)=P(A)+P(B)-P(A ^ B)

V=P(A)*A+P(B)B-P(A ^ B)(A*B)^0.5

As, t=D/V;

t=D/(0.65A+0.35B-P(A ^ B)(A*B)^0.5)

If A and B velocities are independent and there exist event if they are occuring at same time over the any part of time domain, then,

t=D/(0.65A+0.35B-0.2275(A*B)^0.5);

where, P(A ^ B)=P(A)P(B)