Q: A driver of a car travelling at 52km h−1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5s. Another driver going at 3km h−1 in another car applies his brakes slowly and stops in 10s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?

(A) : Initial speed of the car A u=52 kmh−1 =52×185 = 14.44 ms−1v=0t=5With the help of these initial and final points for both the cases, we can plot the graph of speed vs time.Area under the speed-time graph gives the distance covered.u=3kmh−1 =3×185 =0.83Thus car A travels more distance than B.

Question 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?

Accepted Answer

Given :      Diameter of the circular track   D=2R=200  m                  Time     t=2 utes 20 s  =140  sAs the athlete completes one round in 40 seconds, thus he will be at point B after 140  seconds by covering  three and half  rounds.∴  Displacement of the athlete       s′=2R=200 mDistance covered     d=3.5×(2π R)d=3.5×722​×200 =2200   m

Question 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?

Accepted Answer

(a) From A to B,Average velocity=Total displacement/Total timevaverage​=150300​=2m/secso velocity will be 2 m/sec in rightward direction.Average speed=Total distance/time            =300/150=2 m/s(b)From A to C,Average velocity=Total displacement/Total time=200/210=0.952 m/sAverage Speed=totaltimetakentotaldistancetravelled​             =210400​=1.905 m/sec

Question 3

Abdul, while driving to school, computes the average speed for his trip to be 20km h−1. On his return trip along the same route, there is less traffic and the average speed is 30km h−1. What is the average speed for Abdul's trip?

Accepted Answer

Let us assume that Abdul is travelling x kilometer distance while going to school.Distance = Speed x Time1)When it's average speed is 20 km/hrx=v1​×t1​=20×t1​2)When it's average speed is 30 km/hrx=v2​×t2​=30×t2​Vaverage​=totaltimetakentotaldistancetravelled​Vaverage​=v1​x​+v2​x​ x+x​=20x​+30x​ 2x​=24 km h−1

Question 4

A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0m/s2 for 8.0s. How far does the boat travel during this time?

Accepted Answer

We know that Distance is given by second equation of motion , S=ut+21​at2Here S=?=Distanceu=0 = initial velocitya=3m/s2 = acceleratint=8s= timeSo, using the equation we getS=ut+21​at2⇒S=0+21​×3×82⇒S=96m

Question 5

A driver of a car travelling at 52km h−1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5s. Another driver going at 3km h−1 in another car applies his brakes slowly and stops in 10s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?

Accepted Answer

(A)   :         Initial speed of the car A      u=52  kmh−1 =52×185​ = 14.44  ms−1v=0t=5With the help of these initial and final points for both the cases,  we can plot the graph of speed vs time.Area under the speed-time graph gives the distance covered.u=3kmh−1 =3×185​ =0.83Thus car A travels more distance than B.

Question 6

Fig 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?

Accepted Answer

(a) The slope of distance-time graph gives the speed of the objects.            As the slope of object  (B) is the greatest among all, thus  B has the largest speed and hence it is the fastest among all the objects.(b)  :      As the curve of the objects do not intersect with each other even once, hence they cannot be at the same point on the road at the same time.(c)   :      On the y axis  (i.e  distance line), 7 small represents 4 km       ∴  1 small line  =74​ =0.57 km            Position of C at the instant B passes A is  8 km       (d)   ;     Position of B at the instant it passes C is 2nd small line above 4 km i.e 4+2×0.57 =5.14 km         ∴ Distance traveled by B   =5.14−0 =5.14 km

Question 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?

Accepted Answer

Height, s=20mAcceleartion , a=10m/s2By 3rd equation of motion v2=u2+2asv2=0+2×10×20=400⇒ v=20m/sSo, we conclude final velocity , v=20m/sagain, by 1st equation of motion , v=u+at⇒t=av−u​⇒t=1020​=2sSo, it will strike the ground after 2s

Question 8

The speed-time graph for a car is shown in Fig.(a) Find how far does the car travel 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?

Accepted Answer

(a)The total number of divisions on horizontal axis for 2 s is 5.The total number of divisions on vertical axis for 2 m/s is 3.The total displacement of the car during these two seconds will be given as the area under all these squares.The area under the (3×5) squares is given as:A15​=2×2 = 4So, the area under one small square is :A1​=154​Now, the area under the graph for 4 sec is:the number of squares under the graph is :n=57 small squares + 6 half small squaresSo, total number of squares =60.So, the distance traveled by the car is the area under the curve.S=60×154​S=16 m(b)The car moves under the uniform motion after the instant t=6 s because after this instant, the speed of the car becomes constant.

Question 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.

Accepted Answer

(a) Let say a stone is thrown in vertical direction with initial speed u. As the time passes its velocity decreases and at the maximum height it attained, its velocity is zero but it has an acceleration (9.8) m/s2 in the downward direction due to gravity. (b) When a pebble is tied with an end of string and it is whirled in horizontal circular motion with the other end of the string at the centre of circular path, then it has a constant magnitude of centripetal acceleration towards the centre. Hence the acceleration is in perpendicular direction to the motion of pebble.

Question 10

An artificial satellite is moving in a circular orbit of radius 42250km. Calculate its speed if it takes 24 hours to revolve around the earth.

Accepted Answer

Time taken by the satellite to revolve around the earth once      t=24 hours ∴  t=24×3600 s =86400  sRadius of the orbit       R=42250  km =42250000 mLet the velocity of the satellite be   v∴    v=t2πR​=864002×3.14×42250000​⟹ v=3070.95  ms−1  =3.1  kms−1

Question 11

A sphere of radius 0.625 m is rotating about a horizontal axis with angular velocity 2π rad/s and is released from rest in vertical plane under gravity as shown in the figure. The ratio of displacement of point P to that of point C in 0.5 s from the start will be (g=10 m/s2)−

Accepted Answer

Gravity will affect the translation motion of the center of mass. Whereas ω will remain constant throughout the course of motion. During the given span of time the required ratio can be calculated as 2:1.

Question 12

Rectilinear motion of a particle is defined by the equation V=(x2+2x+3)m/s (where V is velocity). The acceleration of the particle when x=1 m is

Accepted Answer

We know that acceleration of the particle is given asa=dtdv​=2xdtdx​+2dtdx​⇒dtdv​=(2x+2)(x2+2x+3)⇒a=4×6=24 ms−2

Question 13

A cyclist is riding with speed of 7.5 m s−1. As he enters a circular turn on a road of radius 80 m, he applies brakes and reduces speed at a rate of 0.5 m/s every second. What is the magnitude of net acceleration of cyclist at this instant?

Accepted Answer

Centripetal acceleration, ac​=rv2​=80(7.5)2​=0.7 m s−2Translational acceleration, at​=0.5 m s−2Net acceleration, a=ac2​+at2​​a=(0.7)2+(0.5)2​=0.86 m s−2

Question 14

An astronaut orbiting the earth in a circular orbit 1000 km above the surface of earth, gently drops a ball out of spaceship. The ball will

Accepted Answer

Ball will move along with the spaceship.  It will be so since gravity will continue to provide the required centripetal acceleration and due to inertia the ball will continue to have the orbital velocity.

Question 15

A particle moves with a constant speed v in a horizontal circular path. The change in its velocity on covering 90∘ will be

Accepted Answer

−−Change in velocity (Δv)=2vsin(2θ​)=2​v

Question 16

In the equation of angular displacement of a particle moving on a circular path is given as θ=2t3+0.5,θ is in radian and t in second. The angular velocity of the particle at t=2 s is:

Accepted Answer

​θ=2t3+0.5ω=dtdθ​=6t2 at t=2 sω=6×(2)2=24 rad s−1​

Question 17

Study the velocity-time graph and calculate.−−(a) The acceleration from A to B(b) The acceleration from B to C(c) The distance covered in the region ABE(d) The average velocity from C to D(e) The distance covered in the region BCFE

Accepted Answer

a. Acceleration from A to B= change in velocity/change in time =25−0/3−0=8.33 m/sb. Acceleration from B to C=18−25/4−3=−7 m/sc. Distance covered in the region ABE=(1/2) (base)(height) =37.5 md. Average velocity from C to D= (initial velocity + final velocity)/2=18−0/2=9 m/se. Distance covered in the region BCFE=(1/2) (sum of parallel sides)(distance between them)=21​(18+25)(4−3)=21.5 m

Question 18

A train is travelling at a speed of 90 kmh−1. Brakes are applied so as to produce a uniform retardation of 0.5 ms−2. How far the train goes before it stops ?

Accepted Answer

Acceleration a=−0.5 m/s2Speed v=90 km/h=25 m/sUsing equation of motion,v=u+at0=25−0.5t25=0.5tt=0.525​t=50 sAgain, using equation of motion,s=ut+21​at2s=25×50−21​×0.5×(50)2s=625 m

Question 19

Use the number line below to answer the following questions:−−Is the initial position positive or negative?Is the final position positive or negative?Is the displacement positive or negative?What is the displacement? (Size and direction (+ or -)

Accepted Answer

a) initial position, xi​=−3 m (negative)b) final position, xf​=−6 m (positive)c) Displacement, xf​−xi​=6−(−3)=9 m (positive)d) Displacement, xf​−xi​=6−(−3)=9 m

Question 20

A teacher enters the examination hall at 8:00 am through the front door and makes 10 rounds along the 15 m wall back and forth. She finally leaves the examination hall at 9:30 am through the same door.
(a) Find the total distance covered by the teacher.
(b) The average speed of her walking is?
(c) What is the displacement of the teacher?
(d) The average velocity of her walk is?
(e) In which condition is the magnitude of average speed is equal to that of average velocity?

Accepted Answer

Total time taken, t=1.5×60=90 min(a) The total distance covered by the teacher, =15×2×10=300 m(b) The average speed of her walking is, =total timetotal distance​=90300​=3.33 m/s(c) The displacement of the teacher, 0 m(d) The average velocity of her walk is, =total timetotal displacement​=900​=0 m/s(e) When the object moves in a single straight line, the magnitude of average velocity equal to average speed.

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