Question

Study the time–distance graph given below for a car to travel to certain places and

answer the following questions:

(i) Calculate the average speed of the car in km/h.

(ii) Between which points is the speed of the car, the greatest?
hi​

Answer 1

Answer :
(a) Graphs shown in fig. , (b) Between 10.40am
to 10.50am
, (c ) 26.5km/h
, (d) 16km/h
, ( e) From E
to F
Solution :
(a) Figure represents the distance - time graph of the car . Time is taken along X - axis and distance is taken along Y - axis.
(b) As the slope of distance - time graph represents speed , therefore , the line with maximum slope will represents maximum speed . As is clear from Fig , line CD
represents maximum speed between 10.40am
to 10.50am
.
( c ) Total distance travelled from 10.05
to 11.40am,s=42km

Total time taken , t=11.40−10.05

=1h35m

=9560h
=
95
60
h

Average speed , vav=total distancetime taken=42km95/60h=26.5km/h
v
a
v
=
total distance
time taken
=
42
k
m
95
/
60
h
=
26.5
k
m
/
h

(d) Between 11.25am
11.25
a
m
and 11.40am
11.40
a
m
,
distance travelled =42−38=4km
=
42
-
38
=
4
k
m

speed=distancetime=4km0.25h=16km/h
speed
=
distance
time
=
4
k
m
0.25
h
=
16
k
m
/
h

(e) Slowing down must give the car minimum speed (=12km/h)
(
=
12
k
m
/
h
)
. This is represented by line EF
E
F
. For this part, distance covered =11.10−11.00=10min=1060h=16h
=
11.10
-
11.00
=
10
min
=
10
60
h
=
1
6
h

speed =distancetime=2km1/6h=12km/h
=
distance
time
=
2
k
m
1
/
6
h
=
12
k
m
/
h
.
Thus the car was forced to slow down to 12km/h
12
k
m
/
h
at distance ranging from 26km
26
k
m
to 28km
28
k
m
.