NCERT Solutions for Class 9th Science Chapter 4 Activity 4.4: Let us calculate — Position-time graphs
Book page 59–60 Updated on2026-09-08
Q1.
In the position-time graph we plotted (Fig. 4.11c), consider a part (say, AB) of the graph as shown in Fig. 4.14. From A, draw a line parallel to X-axis and another line parallel to Y-axis. Repeat the same from B.
Answer
Take A at t₁ = 2 s (position s₁ = 40 m) and B at t₂ = 4 s (position s₂ = 80 m) on the straight line of Fig. 4.11c.
From A, the line parallel to the Y-axis drops to t₁ on the time axis; the line parallel to the X-axis runs across to s₁ on the position axis. Doing the same from B marks off t₂ and s₂.
Dropping perpendiculars from A and B on the position–time line marks off the time interval CA and the change in position BC.
Tip: the two dashed lines from each point are just a way of reading the coordinates of A and B off the axes. They do not represent any motion.
Q2.
Extend the horizontal line from A and a triangle ABC is formed. What do the sides BC and CA of the triangle represent?
Answer
The right-angled triangle ABC has the graph line AB as its hypotenuse. Its two perpendicular sides are:
CA — the horizontal side — represents the change in time, t₂ − t₁.
BC — the vertical side — represents the change in position, s₂ − s₁, which is the displacement between those two instants.
CA = t₂ − t₁ = 4 s − 2 s = 2 s
BC = s₂ − s₁ = 80 m − 40 m = 40 m
Why it happens: CA lies along the direction in which the X-axis measures time, so its length can only be a time interval; BC lies along the direction in which the Y-axis measures position, so its length can only be a change in position. This is why the triangle is worth drawing at all — it converts the abstract "slope" into two sides you can actually measure with the graph's own scales.
Q3.
As per Eq. (4.2a), by dividing the change in position (BC) by the change in time (CA), you get the average velocity.
Answer
Yes — dividing the two sides of the triangle reproduces the definition of average velocity exactly.
average velocity = change in position / time interval [Eq. 4.2a]
here change in position = BC and time interval = CA
v = BC / CA = (s₂ − s₁) / (t₂ − t₁)
Why it happens: BC ÷ CA is, geometrically, the slope of the line AB — how steeply it rises. Physically the same ratio is displacement ÷ time, which is velocity. So "slope of a position–time graph" and "velocity" are not two facts to remember; they are the same ratio read in two languages. In general, the slope of any graph gives the rate of change of the Y-quantity with respect to the X-quantity.
Q4.
By extracting values of time t₁ and t₂, and distances s₁ and s₂ from the graph, the magnitude of average velocity can be calculated.
Answer
Substituting the four values read off Fig. 4.14:
v = (s₂ − s₁) / (t₂ − t₁)
= (80 m − 40 m) / (4 s − 2 s)
= 40 m / 2 s
= 20 m s⁻¹
Because the graph is a straight line, every other pair of points gives the same answer. Check with t₁ = 0 s, t₂ = 6 s:
v = (120 m − 0 m) / (6 s − 0 s) = 120 m / 6 s = 20 m s⁻¹ ✓
Why it happens: a straight line has one slope everywhere. That is exactly what "constant velocity" means, and it is why the choice of the segment AB does not matter here. For a curved position–time graph it would matter: different segments would give different average velocities, because the velocity itself is changing.
Check it yourself: carry the units through the division, as above. Metres divided by seconds gives m s⁻¹ automatically — a useful check that you have not accidentally divided the wrong way round.