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Irodov 1.2Standardkinematicsaverage-velocity

Mean Velocity - Half Distance, Half Time

Problem 1.2
A point traversed half the distance with a velocity v0v_0. The remaining part of the distance was covered with velocity v1v_1 for half the time, and with velocity v2v_2 for the other half of the time. Find the mean velocity of the point averaged over the whole time of motion.
Answer: ⟨v⟩=2v0(v1+v2)2v0+v1+v2\langle v \rangle = \frac{2v_0(v_1 + v_2)}{2v_0 + v_1 + v_2}
Solution Path
In this problem the key was the asymmetry: first half split by distance, second half split by time. Finding each phase time separately and using mean velocity = total distance / total time gives a clean formula where d cancels.
01Problem Restatement
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An object moves along a straight line. It covers the first half of the total distance at one constant speed. Then for the remaining half of the distance, it switches: it moves at a second speed for half the remaining time, and a third speed for the other half of the remaining time.Given:
- v0v_0 (speed for the first half of the distance)
- v1v_1 (speed for the first half of the remaining time)
- v2v_2 (speed for the second half of the remaining time)We need to find: ⟨v⟩\langle v \rangle, the mean velocity averaged over the whole trip, in terms of v0v_0, v1v_1, and v2v_2.
Find ⟨v⟩\langle v \rangle in terms of v0v_0, v1v_1, v2v_2
02Physical Picture and Strategy
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Imagine a delivery truck driving between two cities. For the first half of the route, it drives on a highway at speed v0v_0. Then it exits onto local roads. For the remaining distance, it spends equal amounts of time at two different speeds: v1v_1 through a town, then v2v_2 on a country road.The crucial detail: the first half is split by distance, but the second half is split by time. This asymmetry is what makes the problem interesting.Strategy: Mean velocity equals total distance divided by total time. We will let the total distance be dd, find the time for each phase separately, then combine them.
First half: split by distance. Second half: split by equal time. ⟨v⟩=d/(t1+t2)\langle v \rangle = d / (t_1 + t_2)
03Step 1: Time for Each Phase
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The object covers the first half of the distance (d/2d/2) at constant speed v0v_0. Using the definition of speed (distance = speed times time):t1=d/2v0=d2v0t_1 = \frac{d/2}{v_0} = \frac{d}{2v_0}where t1t_1 is the time for the first half, in the same units as d/v0d/v_0.This means: the faster v0v_0 is, the less time the first half takes.For the second half, the object covers the remaining d/2d/2 in total time t2t_2. It spends t2/2t_2/2 at speed v1v_1 and t2/2t_2/2 at speed v2v_2. The total distance for this phase is:v1⋅t22+v2⋅t22=d2v_1 \cdot \frac{t_2}{2} + v_2 \cdot \frac{t_2}{2} = \frac{d}{2}Factoring out t2/2t_2/2:t22(v1+v2)=d2\frac{t_2}{2}(v_1 + v_2) = \frac{d}{2}t2=dv1+v2t_2 = \frac{d}{v_1 + v_2}This means: the second phase time depends on the sum of the two speeds, not their individual values.
t1=d2v0,t2=dv1+v2t_1 = \dfrac{d}{2v_0}, \quad t_2 = \dfrac{d}{v_1 + v_2}
04Step 2: Assemble the Mean VelocityKEY INSIGHT
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Using the definition of mean velocity (total distance divided by total time):⟨v⟩=dt1+t2\langle v \rangle = \frac{d}{t_1 + t_2}Substituting our expressions for t1t_1 and t2t_2:⟨v⟩=dd2v0+dv1+v2\langle v \rangle = \frac{d}{\dfrac{d}{2v_0} + \dfrac{d}{v_1 + v_2}}This means: the mean velocity is completely determined by v0v_0, v1v_1, and v2v_2. The actual distance dd will cancel out.
⟨v⟩=dd2v0+dv1+v2\langle v \rangle = \dfrac{d}{\dfrac{d}{2v_0} + \dfrac{d}{v_1 + v_2}}
05Step 3: Simplify
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We factor dd out of the denominator, then cancel it with the numerator.⟨v⟩=dd(12v0+1v1+v2)\langle v \rangle = \frac{d}{d\left(\dfrac{1}{2v_0} + \dfrac{1}{v_1 + v_2}\right)}The dd cancels:⟨v⟩=112v0+1v1+v2\langle v \rangle = \frac{1}{\dfrac{1}{2v_0} + \dfrac{1}{v_1 + v_2}}Combining the fractions in the denominator (common denominator 2v0(v1+v2)2v_0(v_1 + v_2)):⟨v⟩=1v1+v2+2v02v0(v1+v2)\langle v \rangle = \frac{1}{\dfrac{v_1 + v_2 + 2v_0}{2v_0(v_1 + v_2)}}Flipping the fraction:⟨v⟩=2v0(v1+v2)2v0+v1+v2\langle v \rangle = \frac{2v_0(v_1 + v_2)}{2v_0 + v_1 + v_2}This means: the mean velocity is a weighted combination of all three speeds. Notice it does not depend on the total distance dd, as expected for a ratio.
⟨v⟩=2v0(v1+v2)2v0+v1+v2\langle v \rangle = \dfrac{2v_0(v_1 + v_2)}{2v_0 + v_1 + v_2}
06Verification and Summary
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Verification: If all three speeds are equal (v0=v1=v2=vv_0 = v_1 = v_2 = v), the object moves at constant speed the entire time, so the mean velocity should just be vv.⟨v⟩=2v⋅2v2v+2v=4v24v=v  ✓\langle v \rangle = \frac{2v \cdot 2v}{2v + 2v} = \frac{4v^2}{4v} = v \; \checkmarkDimensional check:[speed]⋅[speed][speed]=[speed]  ✓\frac{[\text{speed}] \cdot [\text{speed}]}{[\text{speed}]} = [\text{speed}] \; \checkmarkChecks out.Summary: In this problem the key was recognizing that the first half is split by distance and the second half is split by time. We found the time for each phase separately, then used mean velocity equals total distance over total time. The dd cancelled, giving a clean formula in terms of the three speeds.
⟨v⟩=2v0(v1+v2)2v0+v1+v2\boxed{\langle v \rangle = \dfrac{2v_0(v_1 + v_2)}{2v_0 + v_1 + v_2}}
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