AP 9th Class Physical Science Chapter 2 Solution – Motion

AP 9th Class Physical Science Chapter 2 Solution – Motion

Andhra Pradesh SCERT 9th Class Physical Science Chapter 2 Motion Solution for AP 9th Class Physics/Chemistry Exam. Lots of Students of Andhra Pradesh Board will search on internet for Andhra Pradesh Class 9 Physical Science Textbook Solution or Study Material for AP 9th exam. Here you search will end! Here in this page we have provided for all question answer for AP SCERT Chapter 2 Motion.

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AP 9th Class Physics Chapter 2 Motion Solution

(1) As shown in figure23, a point traverses the curved path. Draw the displacement vector from given points A to B (AS5).

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(2) “She moves at a constant speed in a constant direction.” Rephrase the same sentence in fewer words using concepts related to motion.(AS1)

Ans: – If we rephrase the same sentence then it will be “she moves at a constant velocity” or “she moves at zero acceleration”. There is no change in seed so or direction so the velocity must be constant and in constant velocity acceleration is zero.

(3) What is the average speed of a Cheetah that sprints 100m in 4sec.? What if it sprints50m in 2sec? (25 m/s)(AS1, AS7)

Ans: – As we know that average speed = (total distance/total time).

Here given distance= 100m and time= 4sec, so average speed will be (100/4=25) 25m/s.

Another case distance is 50m and time is 2sec so average speed will be (50/2=25) 25m/s.

(4) Correct your friend who says, “ The car rounded the curve at a constant velocity of 70 km/h”.(AS1)

Ans: – As the car rounded the curve at a constant by this direction will be changing at different position so the velocity would not be constant at any more but speed will be constant.

(5) Suppose that the three ball’s shown in figure start simultaneously from the tops of the hills. Which one reaches the bottom first? Explain. See figure 24.(AS2 ,AS1)

Ans: – From figure it is clear that in first case the ball will reach the ground first as it covers the minimum distance comparing to the other with a same speed.

(6) Distance vs time graphs showing motion of two cars A and B are given. Which car moves fast? See figure 25.(AS1)

Ans: – From the s vs t graph we clearly see that the slope of car A is greater than the car B. As in this graph slope denotes the speed so car A will move fast.

(7) Draw the distance vs time graph when the speed of a body increases uniformly.(AS5)

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(8) Draw the distance – time graph when its speed decreases uniformly.(AS5)

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(9) A car travels at a velocity of 80 km/h during the first half of its running time and at 40 km/h during the other half. Find the average speed of the car. (60 km/h) (AS1 , AS7)
 

Ans: – Let the time be t, so the average speed will be (total distance/total time).

The distance cover in first half will be = velocity × t/2 = 80×t/2 = 40t.

The distance covered in other half will be = 40× t/2 = 20t.

So the average speed will be (40t + 20t)/t = 60t/ t = 60km/h.

(10) A car covers half the distance at a speed of 50 km/h and the other half at 40km/h.Find the average speed of the car.(AS1 , AS7)

Ans: – we know the average speed = (total distance / total time).

Let the total distance be D.

So the time required to cover the first half = D/2/ 50 (t=d/v).

Or D/100.

And the time for second half is D/2/40 = D/ 80.

Then total time is (D/100 + D/80)9D/400h.

Total distance is D. So average speed is = D/9D/400 = 400/9 = 44.44km/h.

(11) Derive the equation for uniform accelerated motion for the displacement covered in its nth second of its motion. (Sn = u + a (n – ½ )) (AS1)

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(12) A particle covers 10m in first 5s and 10m in next 3s. Assuming constant acceleration. Find initial speed, acceleration and distance covered in next 2s. (AS1 , AS7)

Ans: –Let the initial speed be u, acceleration be a and distance d.

So according to motion formula S=ut + 1/2at^2.

Here a=? ; S= 10; u=?, t= 5s.

By putting this value we get, 10= 5u + 25/2a…….. (1)

And here t= 3s, speed is (u + 5a) (as v= u + at), s=10,

10 = 3(u + 5a) + 9a/2……. (2)

From 1 and 2 we get,a=1/3 m/s^2 and u=7/6 m/s.

From questions we get distance in 8sec is 20m and distance in 10sec is (s=7/6×10^2 + ½×1/3×10^2= 85/3) 85/3m.

So the distance in next 2s is 85/3 – 20 = 25/3 = 8.33m.

(13) A car starts from rest and travels with uniform acceleration “a” for some time and then with uniform retardation “B” and comes to rest. The time of motion is “t”. Find the maximum velocity attained by it. (AS1, AS7)

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(14) A man is 48m behind a bus which is at rest. The bus starts accelerating at the rate of 1m/s2, at the same time the man starts running with uniform velocity of 10 m/s. What is the minimum time in which the man catches the bus?  (AS1 , AS7)

Ans: – From question we get initial velocity u is o and acceleration is a=1m/s^2.

Let the man catches the bus after time t and at a distance d= 10 × t.

At this time goes up to distance D = 0×t + ½×1×t^2 (by applying s=ut + 1/2at^2).    Or D = t^2/2.

According to question, d – D = 48, or 10t – t^2/2 = 48

Or t = 12s or 8s.

The minimum time required to catch the bus is 8s.

(15) A body leaving a certain point “O” moves with a constant acceleration. At the end ofthe 5 th second its velocity is 1.5 m/s. At the end of the sixth second the body stops andthen begins to move backwards. Find the distance traversed by the body before it stops. Determine the velocity with which the body returns to point “O“? (27m, -9 m/s)(AS1)

Ans: –As we know that v= u + at,

Here given t=5s and v=1.5, so 1.5 = u + 5a…… (1).

And here t=6s, v=0, so 0=u + 6a…… (2).

From equation 1 and 2 we get a=-1.5 m/s^2 and u= 9m/s.

Distance covered in 6s is s= 9×6 + ½× (-1.5) ×6^2(by s= ut + 1/2at^2) =27m.

Now body will go into opposite direction so distance will be negative.

-27= 0 + ½× (-1.5) ×t^2. (In opposite direction u= 0 and a=-1.5,t=?)

So t= 6s.

So the velocity at point O is v = u + at, or 0+ -1.5× 6 = -9m/s.

(16) Distinguish between speed and velocity. (AS1)

Ans: – Speed gives us only magnitude whereas velocity gives us magnitude as well as direction of a body. Speed is scalar but velocity is vector. Rate of change of distance is known as speed and rate of change of displacement is known as velocity.

(17) What do you mean by constant acceleration?(AS1)

Ans: – The constant acceleration ∆v/∆t = constant.

In other ward the rate of change in velocity with respect to time is constant.

(18) When the velocity is constant, can the average velocity over any time interval differ from instantaneous velocity at any instant? If so, give an example; if not explain why.(AS2 , AS1)

Ans: –The average velocity over any time interval can not differ from instantaneous velocity at any instant. As the velocity is constant in every point through out the way and for that average velocity is always equal to constant velocity.

(19) Can the direction of velocity of an object reverse when its acceleration is constant? Ifso give an example; if not, explain why.(AS2 , AS1)

Ans: – The direction of velocity of an object is reverse when its acceleration is constant this statement is valid. For the support of this statement if we throw a ball in the sky after reaching highest position the when it came back its velocity direction is reverse from initial but throughout the way acceleration is same.

(20) A point mass starts moving in a straight line with constant acceleration “a”. At a time t after the beginning of motion, the acceleration changes sign, without change in magnitude. Determine the time t0 from the beginning of the motion in which the point mass returns to the initial position. (AS1)

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 (21) Consider a train which can accelerate with an acceleration of 20cm/s2 and slow down. With deceleration of 100cm/s2.Find the minimum time for the train to travel between the stations 2.7 km apart. (AS1)

Ans: – The distance between the stations is 2.7 km or 270000cm.

Let the distance of acceleration and deceleration is x1 and x2.

So x1 + x2 = 270000….. (1).

From acceleration of motion we get v^2 = u^2 + 2as.

Here u =0 and s= x1, a=20m/s^2, so v^2 = 2ax1 or x1=v^2/2a.

Or x1= v^2/40cm

From deceleration of motion we get v^2= u^2 –2as; here u= 0 and s=x2.   So x2 = v^2/200cm.

So from equation 1 we get, v^2/40 + v^2/200 = 270000, or v = 3000cm/s.

Now the time required in acceleration is, v = u + at, or 3000 = 0 + 20×t, or t = 3000/20= 150s.

And for deceleration time is td = 3000/100= 30s.

So the total time required for the train to travel between the station is (150 + 30 = 180) 180s.

(22) You may have heard the story of the race between the rabbit and tortoise. They started from same point simultaneously with constant speeds. During the journey, rabbit took rest somewhere along the way for a while. But the tortoise moved steadily with lesser speed and reached the finishing point before rabbit. Rabbit awoke and ran, but rabbit realized that the tortoise had won the race. Draw distance vs time graph for this story.(AS5)

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(23) A train of length 50m is moving with a constant speed of 10m/s. Calculate the timetaken by the train to cross an electric pole and a bridge of length 250 m. (AS1)

Ans: –The speed of the train is 10m/s.

When the train will cross the electric pole it cover its own length so the time is 50/10 = 10s.

But when the train cross the bridge it covers its length and the bridge length or (250 + 50 = 300). So the required to cover the bridge is 300/10 = 30s.

(24) Two trains, each having a speed of 30km/h, are headed at each other on the same track. A bird flies off one train to another with a constant speed of 60km/h when they are 60km apart till before they crash. Find the distance covered by the bird and how many trips the bird can make from one train to other before they crash?(AS1)

Ans: –The average speed of the each train is 30km/s.

The distance between two trains is 60 km. And birds speed is 60km/h. So the distance cover by bird in 1 hour is 60×1=60km.

When the two train crash the distance between them will be zero so time to make trips between these two trains tends to zero so infinite no of trips could be made by the birds.

Updated: November 20, 2021 — 12:38 pm

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