1.5: Velocity-time graphs - ppt download - Free Printable

Let's analyze the velocity-time graph provided and answer each question step by step.

---

Understanding the Graph

- The x-axis is time in minutes (min), ranging from 0 to 55 min.
- The y-axis is velocity in meters per minute (m/min).
- The graph shows how velocity changes over time.

We will interpret:

- Positive velocity: motion in the positive direction.
- Negative velocity: motion in the negative direction.
- Zero velocity: at rest.
- Slope of the graph: acceleration.
- Positive slope → positive acceleration.
- Negative slope → negative acceleration.
- Zero slope → constant velocity (zero acceleration).

---

## Answers:

a. Uniform motion in the positive direction?

Uniform motion means constant velocity (straight horizontal line) and positive velocity.

→ Look for a flat segment above the x-axis.

✔ From 10 to 15 minutes, velocity is constant at 60 m/min.

✔️ Answer: [10, 15]

---

b. Uniform motion in the negative direction?

Constant velocity below the x-axis.

→ A flat segment with negative velocity.

Looking at the graph: After t = 40 min, the velocity becomes positive again, but between t = 35 to 40 min, it’s decreasing from 0 to -40, then increases back to 0 at t = 50. But there is no flat segment below zero.

Wait — let's check carefully.

From t = 40 to 50, the velocity goes from -40 to 0, increasing (positive slope), so not uniform.

There is no period where velocity is constant and negative.

✘ So, no interval of uniform motion in the negative direction.

✔️ Answer: None / No such interval

But wait — let’s double-check.

Actually, from t = 40 to 50, the object is accelerating from -40 to 0, so not uniform.

And before that, from t = 30 to 40, velocity goes from 0 to -40 (negative acceleration), but still changing.

So, no horizontal line below zero → no uniform motion in negative direction.

✔️ Answer: None

---

c. Rest?

Rest means velocity = 0.

→ Where the graph crosses or touches the x-axis.

At t = 30 min, velocity = 0.

Also at t = 50 min, velocity = 0.

But only if it's zero for an interval, not just a point.

At t = 30, it's just a point — velocity changes from positive to negative.

At t = 50, same thing.

Is there any interval where velocity is exactly zero?

No — the graph never stays on the x-axis.

So, rest occurs only at points, not intervals.

But usually, "rest" implies zero velocity over an interval.

Since velocity is zero only at isolated points (t = 30, t = 50), not a continuous interval, we say no rest.

However, sometimes "rest" can mean velocity = 0 even momentarily.

But in physics, rest typically means stationary over time.

So unless it's a horizontal line at zero, it's not rest.

But let's see: Is there any horizontal segment at v = 0?

No.

So, no interval of rest.

✔️ Answer: None

---

d. Positive acceleration with positive velocity?

Acceleration = slope of the graph.

Positive acceleration → upward slope.

Positive velocity → above x-axis.

So, look for segments where:
- Slope > 0
- Velocity > 0

→ From t = 0 to 10 min:
- Velocity increases from 0 to 60 → positive slope
- Velocity is positive throughout (0 to 60)

✔ Yes.

✔️ Answer: [0, 10]

---

e. Negative velocity with positive acceleration?

Negative velocity → below x-axis
Positive acceleration → positive slope (upward)

Look for segment where:
- Velocity < 0
- Slope > 0

→ From t = 40 to 50 min:
- Velocity goes from -40 to 0 → increasing → positive slope
- Velocity is negative (from -40 to 0)

Yes!

✔ This matches.

✔️ Answer: [40, 50]

---

f. Negative acceleration with positive velocity?

Negative acceleration → downward slope (negative slope)
Positive velocity → above x-axis

→ From t = 15 to 30 min:
- Velocity decreases from 60 to 0 → negative slope
- Velocity remains positive until t = 30

Yes!

✔ So, [15, 30]

✔️ Answer: [15, 30]

---

g. Negative velocity with negative acceleration?

Negative velocity → below x-axis
Negative acceleration → negative slope (downward)

→ From t = 30 to 40 min:
- Velocity goes from 0 to -40 → decreasing → negative slope
- Velocity is negative (after t = 30)

Yes!

✔ So, [30, 40]

✔️ Answer: [30, 40]

---

h. Zero velocity and negative acceleration?

Zero velocity → at v = 0
Negative acceleration → slope is negative

→ At t = 30 min, velocity = 0
And just before and after: from t = 15 to 30, velocity decreases → negative slope → negative acceleration

So at t = 30, velocity = 0, and acceleration is negative (slope is negative)

✔ Yes, at t = 30

But is it an interval? No — only a point.

But since it's a single instant, we can say at t = 30

✔️ Answer: t = 30

---

i. Positive acceleration and zero velocity?

Positive acceleration → positive slope
Zero velocity → at v = 0

→ At t = 50 min, velocity = 0
And slope is positive (going up from -40 to 0)

So yes, at t = 50, velocity = 0, and acceleration is positive

✔ So, t = 50

✔️ Answer: t = 50

---

j. When does the person change direction?

Change direction when velocity changes sign.

→ From positive to negative: at t = 30 min (crosses from + to -)
→ From negative to positive: at t = 50 min (crosses from - to +)

So, direction changes at t = 30 and t = 50

But t = 30 is the first one.

✔️ Answer: At t = 30 min and t = 50 min

(But often they want the first one — but both are valid.)

But the question says “when” — plural possible.

So: t = 30 and t = 50

✔️ Answer: At t = 30 min and t = 50 min

---

k. What is the distance traveled from [0,10]?

Distance = area under the velocity-time graph (take absolute value of area for distance)

From t = 0 to 10:
- It's a triangle: base = 10 min, height = 60 m/min
- Area = (1/2) × base × height = (1/2) × 10 × 60 = 300 m

All velocity is positive → no need for absolute value

✔️ Answer: 300 meters

---

l. What is the distance traveled from [0,15]?

Break into two parts:
1. [0,10]: triangle → 300 m (as above)
2. [10,15]: rectangle → width = 5 min, height = 60 m/min
→ Area = 5 × 60 = 300 m

Total distance = 300 + 300 = 600 meters

✔️ Answer: 600 meters

---

m. What is the distance traveled from [30,55]?

Note: graph ends at t = 50, so assume [30,50]

From t = 30 to 50:

Break into two parts:

1. [30,40]: triangle below x-axis
- Base = 10 min, height = 40 m/min (but negative)
- Area = (1/2) × 10 × 40 = 200 m (but this is displacement = -200 m)
- Distance = |area| = 200 m

2. [40,50]: triangle above x-axis? Wait — from -40 to 0 → still below zero?

Wait! From t = 40 to 50:
- Velocity goes from -40 to 0 → still negative until t = 50
- So it's a triangle below x-axis, from -40 to 0

Wait — actually, the graph goes from (-40) at t=40 to 0 at t=50 → so it's a triangle below x-axis

But it's increasing (positive slope), but still negative.

So area = (1/2) × base × height = (1/2) × 10 × 40 = 200 m (magnitude)

But since velocity is negative, displacement is -200 m

But distance = total path length = sum of absolute values

So:
- [30,40]: area = (1/2) × 10 × 40 = 200 m (distance)
- [40,50]: area = (1/2) × 10 × 40 = 200 m (distance)

Wait — but from t=40 to 50, velocity goes from -40 to 0 → triangle with base 10, height 40 → area = 200 m

But since it's below x-axis, displacement is -200, but distance is 200 m

So total distance from [30,50] = 200 + 200 = 400 meters

But the question says [30,55] — but graph ends at 50. So likely typo or assume up to 50.

So:

✔️ Answer: 400 meters

---

## ✔ Final Answers:

a. [10, 15]
b. None
c. None
d. [0, 10]
e. [40, 50]
f. [15, 30]
g. [30, 40]
h. t = 30
i. t = 50
j. t = 30 min and t = 50 min
k. 300 m
l. 600 m
m. 400 m

---

Let me know if you'd like a visual explanation or diagram breakdown!