Speed, Time, &Acceleration Graphs. www.gcse.com/fm/stag3.htm. IWBAT:. Interpret & calculate acceleration by reading a speed graph. Speed Time Graphs The first thing to note about these is that, on first glance, they look EXACTLY the same as distance time graphs !
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The first thing to note about these is that, on first glance, they look EXACTLY the same as distance time graphs!
Spot the difference between these two graphs:
The only way you can tell the difference is by reading the labels on the axes.
HINT:Always read the labels on the axes! Fail to do so can mean you completely mis-interpret what it shows.
The speed is increasing, so the object must be accelerating.
The speed is decreasing, so the object must be decelerating. Note that this shape of distance-time graph would mean steady speed back to where you came from. Totally different!
lines at once?
Both the yellow and blue line show increasing speed.
They both reach the same top speed, but the blue one takes longer.
What is the difference? The yellow line shows a greater acceleration.
how far something has travelled. In the example below, a speed of 30 m/s is maintained for 20 seconds:
One way of calculating the distance is to use distance = speed × time. This gives:
distance = 30 × 20 = 600 m
You will already have learned about acceleration. Speed-time graphs can also be used to find accelerations:
The purple line shows an acceleration from 0 to 30 m/s in 20 seconds. (30-0) / 20 = 1.5 m/s/s.
The blue line shows an acceleration from 0 to 20 m/s in 20 seconds. (20-0) / 20 = 1 m/s/s.
You will already have learned about deceleration. We can also find this from speed-time graphs:
The red line shows a deceleration from 30 to 0 m/s in 10 s.
The green line shows a slower deceleration from 10 to 0 m/s over 20 s.
As with accelerations, the gradient (or steepness) of the line indicates the deceleration.
Note that both gradients are negative (they slope downwards) so these are decelerations, not accelerations.
The red line shows an acceleration of 2 m/s/s
The blue line shows a deceleration of 3 m/s/s.
Finally the total distance travelled is: 50 + 200 + 150 = 400 m