Statements:
1. Convert the following:
a. 160 km/h into m/s (Answer: 44.44 m/s)
b. 36 m/s into km/h (Answer: 129.6 km/h)
c. 15 km/h into m/s (Answer: 4.17 m/s)
d. 1 m/s into km/h (Answer: 3.6 km/h)
2. In 10 seconds, a cyclist increases its speed from 5 km/h to 7 km/h, while a car moves from rest to 20 km/h in the same time. Calculate and compare acceleration in each case.
(Answer: 0.055 m/s² and 0.55 m/s²)
3. A ball is thrown straight up such that it took 2 seconds to reach the top, after which it started falling back. What was the speed with which the ball was thrown up?
(Answer: 19.6 m/s)
4. A car is moving with a uniform velocity of 20 m/s for 20 seconds. Then brakes are applied, and it comes to rest with uniform deceleration in 30 seconds. Plot the graph to calculate this distance using a speed-time graph.
(Answer: 1 km)
5. A girl starts her motion on a racing bicycle in a straight line at a speed of 50 km/h. Her speed is changing at a constant rate. If she stops after 60 seconds, what is her acceleration?
(Answer: 0.23 m/s²)
6. Consider the following speed-time graph. Tell:
a. Which part of the graph is showing acceleration, deceleration, and zero acceleration?
b. Calculate the distance covered from 10 seconds to 20 seconds from the graph.
(Answer: (a) OA – Acceleration, BC – Deceleration, AB – Zero Acceleration, (b) 500 m)
1. Convert the following:
a. 160 km/h to m/s:
To convert km/h to m/s:
1 km/h=1000 m/3600 s=5/18 m/s
160 km/h=160×518=44.44 m/s
b. 36 m/s to km/h:
To convert m/s to km/h:
1 m/s=3600 m/1000 m=3.6 km/h
36 m/s=36×3.6=129.6 km/h
c. 15 km/h to m/s:
15 km/h=15×5/18=4.17 m/s
d. 1 m/s to km/h:
1 m/s=1×3.6=3.6 km/h
2. Calculate and compare acceleration:
Given:
- Cyclist’s speed changes from 5 km/h to 7 km/h in 10 seconds.
- Car’s speed changes from 0 km/h (rest) to 20 km/h in 10 seconds.
Convert speeds to m/s:
- Cyclist:
Initial speed=5 km/h =5×1000/3600=1.39 m/s
Final speed=7 km/h=7×1000/3600=1.94 m/s
- Car:
Initial speed=0 m/s
Final speed=20 km/h=20×1000/3600=5.56 m/s\
Using the formula a=Δv/t:
- Cyclist’s acceleration:
- acyclist=1.94−1.39/10=0.055 m/s2
-
Car’s acceleration:
-
acar=5.56−010=0.556 m/s2
3. Speed with which the ball was thrown up:
Using the equation v=u+at, where v=0 m/s (at the top of the throw), is the initial velocity, a=−9.8 m/s2 (acceleration due to gravity), and t=2 seconds (time to reach the top):
0=u+(−9.8)×2
u =
Thus, the speed with which the ball was thrown up is 19.6 m/s.
4. Plot the graph and calculate distance:
For the first 20 seconds, the car moves at a uniform velocity of 20 m/s. Then, it comes to rest in 30 seconds with uniform deceleration.
The total time for the motion is 50 seconds (20 seconds + 30 seconds). The area under the speed-time graph represents the distance traveled.
-
The first 20 seconds:
Distance=Speed×Time=20 m/s×20 s=400 m
-
For the next 30 seconds, the car decelerates uniformly to rest. The average speed during this time is:
Average speed=20+0/2=10 m/s
Thus, the total distance traveled is:
Total distance=400 m+300 m=700 m=0.7 km
5. Girl on a racing bicycle:
Initial speed = 50 km/h = 50×518=13.89 m/s
The girl comes to rest after 60 seconds, so her final speed is 0 m/s.
Using the formula a=v−u/t
a=0−13.89/60=−0.23 m/s2
Thus, her acceleration is 0.23 m/s².
6. Speed-time graph analysis:
a. Acceleration, Deceleration, and Zero Acceleration:
- Acceleration: The part of the graph where the speed increases (positive slope), e.g., from point O to point A (OA).
- Deceleration: The part of the graph where the speed decreases (negative slope), e.g., from point B to point C (BC).
- Zero Acceleration: The part of the graph where the speed remains constant, e.g., from point A to point B (AB).
b. Distance covered from 10 seconds to 20 seconds:
If the graph shows a constant speed during this interval, the distance covered is:
Distance=Speed×Time
Assuming constant speed of 50 m/s between 10 and 20 seconds, then:
Distance=50 m/s×10 s=500 m