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Lesson 8 of 10

How to Find Change in Velocity: Why It's 7, Not 3

速度变化量

Today's question

A ball hits the ground at 5 m/s and bounces up at 2 m/s. Is its change in velocity 3 m/s or 7 m/s?

Lesson notes

A short written version of Lesson 8. The study sheet has more practice questions.

Short answer

The change in velocity is 7 m/s, pointing up, not 3 m/s. Take down as the positive direction: v₀ = +5 m/s and vₜ = −2 m/s, so Δv = −2 − 5 = −7 m/s. The minus sign means up.

Key ideas

  • Change = final − initial, never initial − final. Money from 3000 to 4000 changes by +1000; coins from 100 to 98 change by −2.
  • The change in velocity, Δv, is a vector: an arrow with a size and a direction. Δv = vₜ − v₀, where v₀ is the initial velocity and vₜ is the final velocity.
  • If Δv points the same way as v₀, the object speeds up. If Δv points the opposite way and is shorter than v₀, the object slows down.
  • If Δv points the opposite way to v₀ and its arrow is longer than v₀, the object turns around.
  • On a straight line: choose a positive direction, give v₀ and vₜ their signs, then compute Δv = vₜ − v₀.
  • You can calculate Δv from v₀ and vₜ, but they do not cause it. Something outside the object causes it, such as the brake or the gas pedal.

Step by step

  1. Choose the positive direction. Take down as positive, the direction of the initial velocity.
  2. The ball hits the ground moving down at 5 m/s, so v₀ = +5 m/s.
  3. After the bounce it moves up at 2 m/s. Up is opposite to positive, so vₜ = −2 m/s.
  4. Compute Δv = vₜ − v₀ = −2 − 5 = −7 m/s.
  5. The magnitude is 7 m/s. The minus sign means Δv points up, opposite to the positive direction.
  6. So the change in velocity is 7 m/s upward, not 3 m/s.

Common mistakes

  • Wrong: computing initial − final. Right: always compute final − initial, or the sign comes out backwards.
  • Wrong: subtracting the two speeds, like 5 − 2 = 3. Right: choose a positive direction first, and give each velocity its sign.
  • Wrong: dropping the minus sign, like writing 60 km/h for −60 km/h. Right: a + sign can be dropped, but a − sign cannot.
  • Wrong: thinking a bigger velocity means a bigger Δv. A bullet train is fast, but at constant velocity its Δv = 0.

Practice

Question 1. Take right as positive and find Δv for each motorcycle (in km/h). (a) It speeds up from 30 km/h rightward to 100 km/h. (b) It slows down from 80 km/h rightward to 20 km/h rightward. (c) It brakes and turns around, from 40 km/h rightward to 20 km/h the other way.

Show the answer

(a) +70 km/h, (b) −60 km/h, (c) −60 km/h. Right is positive. (a) Δv = 100 − 30 = 70 km/h. (b) Δv = 20 − 80 = −60 km/h. (c) vₜ = −20, so Δv = −20 − 40 = −60 km/h.

Question 2. A particle moves in a straight line. The magnitude of its initial velocity is 2 m/s and the magnitude of its final velocity is 4 m/s. Then ( ) (more than one may be correct)
A. The magnitude of the change in velocity may be 6 m/s
B. The magnitude of the change in velocity may be 4 m/s
C. The change in velocity may point the same way as the initial velocity
D. The change in velocity may point opposite to the initial velocity

Show the answer

A, C and D. Take the initial direction as positive, so v₀ = +2 m/s. If vₜ = +4 m/s, Δv = 4 − 2 = +2 m/s (C). If vₜ = −4 m/s, Δv = −4 − 2 = −6 m/s (A, D). A Δv of magnitude 4 m/s never happens (B is wrong).

Words in English and Chinese

change in velocity (Δv)
速度变化量
initial velocity
初速度
final velocity
末速度
positive direction
正方向
vector
矢量
scalar
标量
magnitude
大小

Sum it up before you move on.

Write this lesson in one line:

  1. What was the question?
  2. What is the simple version?
  3. What would you have got wrong?
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