Question

22. Figure 2.E6 shows the speed-time graph of a bus.
Speed (km/h)
10
20.
30
Time (minutes)
Fig. 2.E6
(a) In which period is the bus accelerating?
(b) In which period is the bus decelerating?
(c) What is the distance covered during its
acceleration?
(d) What is the distance covered during its
deceleration?
(e) What is the average speed in the entire
journey?​

Answer 1

Given:

A speed - time graph has been provided as follows.

To find:

(a) In which period is the bus accelerating?

(b) In which period is the bus decelerating?

(c) What is the distance covered during its

acceleration?

(d) What is the distance covered during its deceleration?

(e) What is the average speed in the entire journey?

Calculation:

Part a)

The bus is accelerating in the 1st part of the graph where the slope is positive . Part A-B of graph shows Accelerated Motion.

Part b)

The bus is decelerating in the second part of the graph where the slope is negative. Part B-C of graph shows decelerated motion.

Part c)

Distance covered during acceleration is :

$\sf{d = area \: of \: triangle \: ABO}$

$\sf{ = > d = \dfrac{1}{2} \times 10 \times 40 }$

$\sf{ = > d = 200 \: m }$

Part d)

Distance cover during deceleration is :

$\sf{d = area \: of \: triangle \: BOC}$

$\sf{ = > d = \dfrac{1}{2} \times (20 - 10) \times 40 }$

$\sf{ = > d = \dfrac{1}{2} \times 10 \times 40 }$

$\sf{ = > d = 200 \: m}$

Part e)

Average speed in the entire journey is :

$\sf{avg. \: v = \dfrac{total \: distance}{total \: time} }$

$\sf{ = > avg. \: v = \dfrac{200 + 200}{10 + 10} }$

$\sf{ = > avg. \: v = \dfrac{400}{20} }$

$\sf{ = > avg. \: v = 20 \: m {s}^{ - 1} }$

Answer 2

Answer:

(a) 0-10 minute

(b) 0- 20 minute

(c) 10/3 km

(d) 10/3 km

(e) 20 km/h