MOTION
EXAMPLE 8.7
QUESTION : the brakes applied to a car produxe an acxeration of 6m s in the opposite direction to the motion .if the car takes takes 2 s to stop after the application of brakes ccalculate the distance it travlea during thia time.
ANSWER : a : 6 m s , t : 2 s , v : 0m s
v = u +at
0 = u + (-6 s) * 2 s
or u = 12 m s
s= ut + 1/2at2
= 12 ms * 2ms +1/2 (6ms) (2ms)
=24 m - 12 m
= 12 m
EXAMPLE 8.6
QUESTION : a car accekerates uniformoly from 18 km / h to 36 km / h in 5 sec.cakculate the accekeration nd the distance covered by the var in tgat time
answer. u = 18 km / h
= 5 m / sec
v = 36 km / h
= 10 m / sec
t = 5 sec
(i) a = u-v
---
t
= 10 m s - 5m s
-------------
5s
= 1 m s
(ii) s = ut + 1/2 at2
= 5 m s * 5 s * 1/2 * 1ms * 5 s
= 25 m + 12.5 m
= 37.5 m
Excercise
1. An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?
Answer
Diameter of circular track (D) = 200 m
Radius of circular track (r) = 200 / 2=100 m
Time taken by the athlete for one round (t) = 40 s
Distance covered by athlete in one round (s) = 2π r
= 2 x ( 22 / 7 ) x 100
Speed of the athlete (v) = Distance / Time
= (2 x 2200) / (7 x 40)
= 4400 / 7 × 40
Therefore, Distance covered in 140 s = Speed (s) × Time(t)
= 4400 / (7 x 40) x (2 x 60 + 20)
= 4400 / ( 7 x 40) x 140
= 4400 x 140 /7 x 40
= 2200 m
Number of round in 40 s =1 round
Number of round in 140 s =140/40
=3 1/2
After taking start from position X,the athlete will be at postion Y after 3 1/2 rounds as shown in figure
Hence, Displacement of the athlete with respect to initial position at x= xy
= Diameter of circular track
= 200 m
2. Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph's average speeds and velocities in jogging (a) from A to B and (b) from A to C?
Answer
Total Distance covered from AB = 300 m
Total time taken = 2 x 60 + 30 s
=150 s
Therefore, Average Speed from AB = Total Distance / Total Time
=300 / 150 m s-1
=2 m s-1
Therefore, Velocity from AB =Displacement AB / Time = 300 / 150 m s-1
=2 m s-1
Total Distance covered from AC =AB + BC
=300 + 200 m
Total time taken from A to C = Time taken for AB + Time taken for BC
= (2 x 60+30)+60 s
= 210 s
Therefore, Average Speed from AC = Total Distance /Total Time
= 400 /210 m s-1
= 1.904 m s-1
Displacement (S) from A to C = AB - BC
= 300-100 m
= 200 m
Time (t) taken for displacement from AC = 210 s
Therefore, Velocity from AC = Displacement (s) / Time(t)
= 200 / 210 m s-1
= 0.952 m s-1
3. Abdul, while driving to school, computes the average speed for his trip to be 20 km h−1. On his return trip along the same route, there is less traffic and the average speed is 40 km h−1. What is the average speed for Abdul’s trip?
Answer
The distance Abdul commutes while driving from Home to School = S
Let us assume time taken by Abdul to commutes this distance = t1
Distance Abdul commutes while driving from School to Home = S
Let us assume time taken by Abdul to commutes this distance = t2
Average speed from home to school v1av = 20 km h-1
Average speed from school to home v2av = 30 km h-1
Also we know Time taken form Home to School t1 =S / v1av
Similarly Time taken form School to Home t2 =S/v2av
Total distance from home to school and backward = 2 S
Total time taken from home to school and backward (T) = S/20+ S/30
Therefore, Average speed (Vav) for covering total distance (2S) = Total Distance/Total Time
= 2S / (S/20 +S/30)
= 2S / [(30S+20S)/600]
= 1200S / 50S
= 24 kmh-1
4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s−2 for 8.0 s. How far does the boat travel during this time?
Answer
Given Initial velocity of motorboat, u = 0
Acceleration of motorboat, a = 3.0 m s-2
Time under consideration, t = 8.0 s
We know that Distance, s = ut + (1/2)at2
Therefore, The distance travel by motorboat = 0 x 8 + (1/2)3.0 x 8 2
= (1/2) x 3 x 8 x 8 m
= 96 m
5
Answer
As given in the figure below PR and SQ are the Speed-time graph for given two cars with initial speeds 52 kmh-1 and 3 kmh-1 respectively.
Distance Travelled by first car before coming to rest =Area of △ OPR
= (1/2) x OR x OP
= (1/2) x 5 s x 52 kmh-1
= (1/2) x 5 x (52 x 1000) / 3600) m
= (1/2) x 5x (130 / 9) m
= 325 / 9 m
= 36.11 m
Distance Travelled by second car before coming to rest =Area of △ OSQ
= (1/2) x OQ x OS
= (1/2) x 10 s x 3 kmh-1
= (1/2) x 10 x (3 x 1000) / 3600) m
= (1/2) x 10 x (5/6) m
= 5 x (5/6) m
= 25/6 m
= 4.16 m
6. Fig 8.11 shows the distance-time graph of three objects A, B and C. Study the graph and answer the following questions:
(a) Which of the three is travelling the fastest?
(b) Are all three ever at the same point on the road?
(c) How far has C travelled when B passes A?
(d)How far has B travelled by the time it passes C?
Answer
(a) Object B
(b) No
(c) 5.714 km
(d) 5.143 km
Therefore, Speed = slope of the graph
Since slope of object B is greater than objects A and C, it is travelling the fastest.
(b) All three objects A, B and C never meet at a single point. Thus, they were never at the same point on road.
On the distance axis:
7 small boxes = 4 km
Therefore,1 small box = 4 / 7 Km
Initially, object C is 4 blocks away from the origin.
Therefore, Initial distance of object C from origin = 16 / 7 Km
Distance of object C from origin when B passes A = 8 km
Distance covered by C
Page No: 113
7. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s−2, with what velocity will it strike the ground? After what time will it strike the ground?
Answer
Let us assume, the final velocity with which ball will strike the ground be 'v' and time it takes to strike the ground be 't'
Initial Velocity of ball, u =0
Distance or height of fall, s =20 m
Downward acceleration, a =10 m s-2
As we know, 2as =v2-u2
v2 = 2as+ u2
= 2 x 10 x 20 + 0
= 400
∴ Final velocity of ball, v = 20 ms-1
t = (v-u)/a
∴Time taken by the ball to strike = (20-0)/10
= 20/10
= 2 seconds
8. The speed-time graph for a car is shown is Fig. 8.12.
(a) Find out how far the car travels in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.
(b) Which part of the graph represents uniform motion of the car?
Answer
(a)
The shaded area which is equal to 1 / 2 x 4 x 6 = 12 m represents the distance travelled by the car in the first 4 s.
(b)
The part of the graph in red colour between time 6 s to 10 s represents uniform motion of the car.
9. State which of the following situations are possible and give an example for each of these:
(a) an object with a constant acceleration but with zero velocity.
(b) an object moving in a certain direction with an acceleration in the perpendicular direction.
Answer
(a) Possible
When a ball is thrown up at maximum height, it has zero velocity, although it will have constant acceleration due to gravity, which is equal to 9.8 m/s2.
(b) Possible
When a car is moving in a circular track, its acceleration is perpendicular to its direction.
10. An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.
Answer
Radius of the circular orbit, r= 42250 km
Time taken to revolve around the earth, t= 24 h
Speed of a circular moving object, v= (2π r)/t
=[2× (22/7)×42250 × 1000] / (24 × 60 × 60)
=(2×22×42250×1000) / (7 ×24 × 60 × 60) m s-1
=3073.74 m s -1
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