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The graph below represents the relationship between speed and time for a car moving in a straight line.

What is the acceleration of the car?

a
0.10 m/s2

b
1.0 m/s2

c
0 m/s2

d
10 m/s2

3 answers

To determine the acceleration of the car from a speed-time graph, you need to look at the slope of the graph. The slope of a speed vs. time graph represents acceleration, which can be calculated as:

\[ \text{Acceleration} (a) = \frac{\Delta \text{Speed} (\Delta v)}{\Delta \text{Time} (\Delta t)} \]

Without the visual of the graph, I would assume you may have specific points in speed and time that you can use to calculate this slope (change in speed over change in time). If you have two points on the line, find their coordinates and plug them into the formula.

Here's how to do it:

  1. Identify two distinct points on the line of the graph. For example:

    • Point 1: (time1, speed1)
    • Point 2: (time2, speed2)

  2. Calculate the change in speed (\(\Delta v\)): \[ \Delta v = \text{speed2} - \text{speed1} \]

  3. Calculate the change in time (\(\Delta t\)): \[ \Delta t = \text{time2} - \text{time1} \]

  4. Divide the change in speed by the change in time to get the acceleration: \[ a = \frac{\Delta v}{\Delta t} \]

If the graph shows a straight horizontal line, then the speed is constant, and the acceleration would be 0 m/s². If you find a positive slope, it would indicate positive acceleration, and a negative slope would indicate deceleration.

Please check the graph and apply the calculations as described. If you provide specific numbers, I can help calculate it directly. Otherwise, choose the answer from the options based on your slope calculation:

  • 0.10 m/s²
  • 1.0 m/s²
  • 0 m/s²
  • 10 m/s²
The graph shows a straight line going diagonally upward with a time of 0 seconds and 0 m/s to 3.0 seconds at 30m/s

Given that the graph shows a straight line going diagonally upward from a speed of 0 m/s at 0 seconds to a speed of 30 m/s at 3.0 seconds, we can calculate the acceleration using the formula for acceleration:

\[ a = \frac{\Delta v}{\Delta t} \]

Where:

  • \(\Delta v\) is the change in speed
  • \(\Delta t\) is the change in time

From the data you provided:

  • Initial speed (\(v_1\)) = 0 m/s at \(t_1 = 0\) seconds
  • Final speed (\(v_2\)) = 30 m/s at \(t_2 = 3.0\) seconds

Now, we can find \(\Delta v\) and \(\Delta t\):

  1. Change in speed (\(\Delta v\)): \[ \Delta v = v_2 - v_1 = 30 , \text{m/s} - 0 , \text{m/s} = 30 , \text{m/s} \]

  2. Change in time (\(\Delta t\)): \[ \Delta t = t_2 - t_1 = 3.0 , \text{s} - 0 , \text{s} = 3.0 , \text{s} \]

Now, substitute these values into the acceleration formula:

\[ a = \frac{\Delta v}{\Delta t} = \frac{30 , \text{m/s}}{3.0 , \text{s}} = 10 , \text{m/s}^2 \]

So, the acceleration of the car is 10 m/s².

The correct answer is d) 10 m/s².