NCERT Solutions for Class 9th Science Chapter 4 In-text Questions — Velocity-time graphs

Book page 61–62 Updated on2026-09-08

Q1.

What does the shape of the velocity-time graph indicate about the nature of motion?

Answer

Here the slope is the acceleration, so the shape reports on how the velocity is changing.

Shape of the velocity–time graph	What it means	Book's example
Straight line parallel to the time axis	Velocity constant, acceleration zero	Fig. 4.17a — car at a steady 20 m s⁻¹
Straight line sloping upward	Velocity increasing at a constant rate; acceleration constant and along the velocity	Fig. 4.17b — 0 to 15 m s⁻¹ in 30 s
Straight line sloping downward	Velocity decreasing at a constant rate; acceleration constant and opposite to the velocity	Fig. 4.17c — 15 m s⁻¹ down to 0 in 30 s
Curved line	Acceleration itself is changing	Not treated in this chapter

Table 4.5 shows why Fig. 4.17b is straight: the velocity rises by exactly 2.5 m s⁻¹ in every 5 s interval — equal changes in equal times.

a = (2.5 m s⁻¹ − 0 m s⁻¹) / (5 s − 0 s) = 0.5 m s⁻²
a = (5.0 m s⁻¹ − 2.5 m s⁻¹) / (10 s − 5 s) = 0.5 m s⁻² … the same throughout

Why it happens: acceleration is change in velocity ÷ change in time, which on these axes is rise ÷ run — the slope. A fixed slope therefore means a fixed acceleration, and a zero slope means no acceleration at all, however large the velocity may be.

Q2.

Which physical quantities can be obtained from a velocity-time graph?

Answer

Three, and the third one is the reason this graph is so useful:

The velocity at any instant — read the Y-value directly.
The average acceleration — the slope of the line between two points.
The displacement — the area enclosed between the line and the time axis for that interval.

slope: a = BC / CA = (v − u) / (t₂ − t₁)
from Fig. 4.17d between 10 s and 20 s:
a = (10 m s⁻¹ − 5 m s⁻¹) / (20 s − 10 s) = 5 m s⁻¹ / 10 s = 0.5 m s⁻²

area: displacement between 0 s and 6 s in Fig. 4.18a
= area of rectangle OABC = OA × OC
= 20 m s⁻¹ × 6 s = 120 m

Why it happens: the area is a product of a height (velocity, in m s⁻¹) and a width (time, in s), and m s⁻¹ × s = m — a length. That is not a coincidence: for constant velocity, displacement = velocity × time is exactly the area of the rectangle. For a sloping line the same idea holds; you just split the shape into a rectangle and a triangle.

Q3.

Can you calculate some other physical quantity from the velocity-time graph?

Answer

Yes — the displacement, from the area enclosed between the graph line and the time axis.

For the accelerating car of Fig. 4.18b, the displacement between 10 s and 20 s is the area of the shape ABDE, which splits into a rectangle and a triangle:

displacement = area of ABDE
= area of rectangle ACDE + area of triangle ABC
= (CD × DE) + (½ × CA × BC)
= (5 m s⁻¹ × 10 s) + (½ × 10 s × 5 m s⁻¹)
= 50 m + 25 m = 75 m

Displacement between 10 s and 20 s = rectangle (50 m) + triangle (25 m) = 75 m.

Why it happens: the rectangle is the distance the car would have covered had it kept its starting velocity of 5 m s⁻¹ for the full 10 s. The triangle is the extra distance it gains because it is speeding up. Adding them is just adding "what it would have done" to "what the acceleration added".

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