P5 Distance-Time and Velocity-Time Graphs (page 209) Flashcards
If an object moves in a straight line, its distance travelled can be plotted on a distance-time graph.
On graph 1 page 209, describe the gradient = speed?
1) Gradient = speed. (The steeper the graph, the faster it’s going).
this is because : speed = distance ÷ time = (change in vertical axis) ÷ (change in horizontal axis)
On graph 1 - page 209, explain what the flat sections is?
Flat sections are where it’s stationary - it’s stopped.
On graph 1 - page 209, explain what the stright uphill sections mean?
they mean it is travelling at a steady speed.
On graph 1 - page 209, explain what the curves represent?
they represent acceleration or decelleration (p.208).
On graph 1 - page 209, explain what a steepening curve means?
it’s speeding up (increasing gradient).
On graph 1 - page 209, explain what levelling off curve means?
a levelling off curve means it’s slowing down.
On graph 1 - page 209, explain what a steepening curve means?
i’s speed up (increasing gradient).
On graph 1 - page 209, explain wha levelling off curve means?
it’s slowing down
On graph 1 - page 209, if the object is changing speed (accelerating) how can you find its speed at a point?
by finding the gradient of the tangent to the curve at that point, see page 146.
How an object’s velocity changes as it travels can be plotted on a velocity-time graph.
On graph 2 - page 209, explain what Gradient = acceleration means?
since accerlation is change in velocity ÷ time.
On graph 2 - page 209, explain what the flat sections represent?
they represent travelling at a steady speed.
On graph 2 - page 209, explain what the steeper graph line is?
the steeper the graph, the greater the acceleration or deceleration.
On graph 2 - page 209, explain the uphill sections (/) mean
acceleration
On graph 2 - page 209, explain the downhill sections () mean?
deceleration
On graph 2 - page 209, explain what a curve means?
changing acceleration
(if the graph is curved you can use a tangent to curve at a point to find the acceleration at that point).
On graph 2 - page 209, explain the area under any section of the graph (or all of it) is equal to?
equal to the distance travelled in that time interval.
On graph 2 - page 209, explain if the section under the graph is irregular, and how to fid the area?
if its irregular, its easier to find the area by counting the squares under the line and multiplying the number by the value of one square.
The velocity-time graph of a car’s journey is plotted. see diagram 3 on page 209.
a) calculate the acceleration of the car over the first 10 s
b) How far does the car travel in the first 15 s of the journey?
a) this is just the gradient of the line.
a = Δv ÷ t = 20 ÷ 10 = 2 m/s²
b) Split the area into a triangle and a rectangle, then add together their areas.
Or find the value of one square, count the total number of squares under the line, and then multiply these two values together.
Area = (½ x 10 x 20) + (5 x 20)
= 200 m
1 square = 2 m/s x s = 2 m
Area = 100 squares = 100 x 2 = 200 m
Make sure you know the difference between distance-time and velocity-time graphs, and how to interpret them.
Sketch the distance-time graph for an object that accelerates before travelling at a steady speed. (2 marks).
see graph on page 247 p.209 Q1
1 mark for a curved line with an increasing positive gradient. 1 mark for the line becoming a straight line with a positive gradient.
A stationary car starts accelerating increasingly for 10 s until it reaches a speed of 20 m/s. It travels at this speed for 0 s until the driver sees a hazard and breakes. He decelerates uniformly, coming to a stop 4 s after breaking. Draw a velocity-time graph for this journey (3 marks)
See graph on page 274 p.209 Q2
1 mark for an upwards curved acceleration line to 20 m/s. 1 mark for a straight line representing steady speed. 1 mark for a straight line representing deceleration.
