Graphing Speedvs Time-1 - Name: Date: ______ Period: - Studocu - Free Printable

Explanation:
Let’s go step by step using the data:

| Time (s) | Speed (km/hr) |
|----------|----------------|
| 0 | 0.0 |
| 2 | 10.0 |
| 4 | 20.0 |
| 6 | 30.0 |
| 8 | 40.0 |
| 10 | 50.0 |

We can see that for every 2 seconds, speed increases by 10 km/hr.

So the change in speed per second is:
10 km/hr ÷ 2 s = 5 km/hr per second
→ This is the acceleration.

But note: units are mixed — time is in seconds, speed is in km/hr. For physics, we’d usually convert to m/s, but since all questions use km/hr and seconds consistently, and the pattern is linear, we can treat it as a proportional relationship *within this problem*.

Let’s answer each question:

1. As time increases, what happens to the speed?
From the table: speed goes from 0 → 10 → 20 → 30 → 40 → 50 as time goes 0 → 2 → 4 → 6 → 8 → 10.
So speed increases steadily (linearly) with time.

2. What is the speed at 5 s?
At t = 4 s, speed = 20.0 km/hr
At t = 6 s, speed = 30.0 km/hr
Since it’s linear, halfway between 4 s and 6 s (i.e., at 5 s), speed is halfway between 20 and 30:
(20 + 30)/2 = 25.0 km/hr

3. Assuming constant acceleration, what would be the speed at 14 km/hr?
Wait — careful! The question says: *“what would be the speed at 14 km/hr?”* That doesn’t make sense — speed is *in* km/hr. Likely a typo — it should be:
“what would be the speed at 14 s?”
Let’s check context: previous question asks speed at 5 s; next asks time for 45 km/hr — so yes, this must mean at t = 14 s.

We know:
- At t = 0, v = 0
- Acceleration = Δv / Δt = (10 km/hr) / (2 s) = 5 km/hr per second
So v = a × t = 5 × t (with v in km/hr, t in seconds)

At t = 14 s:
v = 5 × 14 = 70.0 km/hr

✔ Check: at t = 10 s, v = 5 × 10 = 50 → matches table. Good.

4. At what time would the object reach a speed of 45 km/hr?
Use v = 5t → t = v / 5 = 45 / 5 = 9 seconds

Check: at t = 8 s, v = 40; t = 10 s, v = 50 → 45 is halfway → t = 9 s. Correct.

5. What is the object’s acceleration?
Acceleration = change in speed / change in time
Use any pair: from t = 0 to t = 10 s:
Δv = 50.0 − 0.0 = 50.0 km/hr
Δt = 10 − 0 = 10 s
a = 50.0 / 10 = 5.0 km/hr per second

(Note: In SI units, this would be converted, but the problem uses km/hr and seconds, and all answers expect this unit.)

6. What would the shape of the graph be if a speed of 50.0 km/hr is maintained from 10 s to 20 s?
From 0–10 s: speed increases linearly → straight line upward.
From 10–20 s: speed stays constant at 50.0 → horizontal line.
So overall: a straight line rising to (10, 50), then a flat horizontal line to (20, 50).
Shape: a line with positive slope, then a horizontal line — i.e., a “corner” or “kink” at t = 10 s.

Final Answer:
1. The speed increases steadily (linearly).
2. 25.0 km/hr
3. 70.0 km/hr
4. 9 s
5. 5.0 km/hr per second
6. A straight upward-sloping line from (0,0) to (10,50), then a horizontal line from (10,50) to (20,50)