Chapter 4
Speed and time
Exercise
1) The following motions are classified as simple linear, reciprocating or oscillatory motion.
A) The speed of your hand at a distance.
Answer: Oscillation speed.
b) It is the speed of the bagi tow on the road.
Answer: Simple linear.
C) The speed of the child standing up on the Merry-go-Round.
Answer: Circular.
D) The position of the child standing on the see-saw is its speed.
Answer: Oscillation speed.
(e) Speed of electric clock hands.
Answer: Oscillation speed.
f) Now the speed of Pon Dalanga train.
Answer: Simple linear motion.
2) Which of the following statements is not true?
a) Fundamental unit of time
Answer: Pure.
b) Every object moves faster.
Answer: Pure.
c) The distance between two faces is measured in kilometers.
Answer: Pure.
d) It is not constant given the period of the oscillator.
Answer: Impure.
e) Now train speed is expressed in m/h.
Answer: Pure.
3) It swings 20 times in simple swing and takes 32 seconds. Determine the period of the oscillator.
Answer: It takes 32 seconds to swing 20 times.
The period of oscillation is - 32
= 1.6 seconds.
4) The distance between the two shrines is 240 km. Now the train takes 4 hours to cover this distance. Determine the speed of the train.
Answer : Given
Distance between the two stations = 240 km.
Time = 4 hours
I know, speed = distance/time.
This gives the speed = 240/4 km/h.
= 60 km/h.
5) Looking at the time before 08.30 hrs the clock now shows the distance of 57321.0 km. If before 08.50 the reading of distance meter changes to 57336.0 km. Determine the distance traveled by the tent car. Display the speed of the car in km/h units within this time.
Answer : Given
Od'meter original reading = 57321.0 km. Odometer's last reading = 57336.0 km.
Early reading of clock = 08:30 AM
Last reading of clock = 08:50 AM
Etiya, Muth Distance = (57336.0 –57321.0) km. = 15.0 km.
Minutes = (08.50 – 08.30) minutes = 20 minutes
This determines the speed = 15/20 km/min.
= 0.75 km/min
1 hour = 60 minutes
Speed in km/h = 0.75 x 60 = 45 km/h.
6) It takes 15 minutes to go to school from Chalmai Chaikelere house. If the speed of Chaikkelkhan is 2 m./s. Yes, the distance from the house to the school is determined.
Answer : Given
Speed per cycle = 2 m. / is
Time = 15 minutes
= 15 x 60,
= 900 ch
Distance covered = speed x time
= 2 m. / sq., x 900 sq.,
= 1800 m.
Distance from Chalma house to school = 1.800 km.
7) Observe the shape of the speed distance-time graph for the conditions mentioned below-
(a) The car is now moving at constant speed
(b) The road is now occupied by cars
8) Which of the following relations is pure?
