Ix physics motion_and_rest_solved_numerical

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IX Physics Motion and Rest
CBSE chapter-wise MCQ Multiple Choice Questions, Test Paper, Sample paper on CCE pattern for class 9 science
Motion. Distance and displacement, velocity; uniform and non-uniform motion along a straight line; acceleration,
distance-time and velocity-time graphs for uniform motion and uniformly accelerated motion, equations of motion
by graphical method; elementary idea of uniform circular motion.
IX Physics Motion and Rest solved Numerical:
. in a circular track (distance 400 m) an athlete runs 1/4 the of the ground. So what would be the displacement?
Ans: Given length of the circular track = 400m. Since the athlete runs ¼ of the circular track
the distance covered by the athlete is 100m. The displacement is the shortest distance
between the initial and final position, therefore, displacement in above question is length
of the straight line AB as given in the diagram below.
To calculate length of the line AB we need to first calculate the radius of the circular track.
Now, Length of track = 2 R 400 = 2 R R =
Now, In right OAB; AB2
= OA2
+ OB2
AB2
= R2
+ R2
= =
Q. A train travels 40 km at a uniform speed of 30 km/hr. Its average speed after travelling another 40 km is 45 km/hr
for the whole journey. It's speed in the second half of the journey is ?
Ans: Suppose speed of the train in second half be v km/hr ; Total distance travelled by the train = 80 km
Total time taken T =
Time taken to cover first half, T1 = = hrs
Time taken to cover second half , T2 =
Now, T1 + T2 = T
Thus the speed to travel second-half = 90km/hr
Q. a motorcyclist drives from a to b with the uniform speed of 30 km/h-1 and returns back with the speed of 20
km/h-1.find the average speed?

Given: Speed to travel from point a to b = 30km/hr; Speed of return journey = 20 km/hr
Let distance between a and b is x km so, Total distance covered = 2x km
Total time = t1 +t2 =
Average speed = Total distance/ total time =
Q. A cheetah is the fastest land animal and it can achieve a peak velocity of 100 km per hour upto distances less than
500 metres. If the cheetah spots his prey at a distance of 100 metres what is the minimum time it will take to get its
prey?
Ans: If the cheetah spots the prey at its top speed, the cheetah will hunt down the prey with the speed of 100 Km/h.=
27.7 m/s Now, speed = d/t time = d/s= 100/27.7 = 3.6 sec.
So, the minimum time the cheetah will take to get the prey is 3.6 s.
Q. A police jeep is chasing with velocity of 45 km/h. A thief in another jeep moving with a velocity of 153 km/h. police
fires a bullet with muzzle velocity of 180 m/s. the velocity it will strike the car of the thief is ___?
Ans: Given: Velocity of police jeep =45km/hr = 12.5 m/s Velocity of thief’s jeep = 153 km/hr = 42.5 m/s
Velocity of bullet = 180 m/s
Now, since bullet is fired from police jeep which is going at 42.5 m/s,
Therefore, velocity of bullet is (180+12.5) = 192.5m/s.
To calculate velocity with which bullet will hit the thief we use concept of relative velocity.
Therefore, we have VBT = VB - VT Here VB is velocity of bullet and VT is velocity of thief
VBT = 192.5m/s - 42.5 m/s = 150m/s
Q. A car moves with a speed of 30km/h for half an hour; 25km/h for 1 hr and 40km/hr for 2 hrs. Find average speed.
Ans: Distance travelled with a speed of 30km/h for half an hour = 30km/h x1/2hr =15km
Distance travelled with a speed of 25km/h for an hour = 25km/h x1hr =25km
Distance travelled with a speed of 40km/h for 2 hour = 40km/h x2hr =80km
Average speed =

Q. If a car travels first 40km at speed of 20km/h and next 80km at a speed of 40km/h. What is the average speed of
cat during journey?
Ans: Average speed =
Q. On a 120 km long track, a train travels the first 30 km at a speed of 30km/h .How fast train travel the next 90km so
that average speed would be 60km/h?
Ans: Let train travel the next 90km at speed of x km/h, Average speed given =60km/h
Average speed =
60km/h = 60x + 540 = 120x 540/60=x x=90km/hr
Q. Q. An object covers half the distance with the speed of 20m/s and other half with a speed of 30m/s.Find average
speed?
Ans: Let S be the total distance traveled by the object.
It covers (S/2) with speed 20 m/s. Time taken to cover this (S/2) is = (S/2)/20 = S/40
It covers the other (S/2) with speed 30 m/s. Time taken to cover this (S/2) is = (S/2)/30 = S/60
Average speed = Total distance/Total time = (S/2 + S/2)/(S/40 + S/60) = 24 m/s
Q. A 150 metres long train crosses a man walking at the speed of 6km/h in the opposite direction in 6 seconds. The
speed of the train in km/h is how much?
Ans: Let 'v' be the speed of the train with respect to the ground. The velocity of the man with respect to ground is (- 6
) km/h (negative sign is included because the man and the train are moving in opposite directions).
So, the velocity of the man with respect to the train (relative speed)= [v - (-6)] = (v + 6) km/h
Time taken by the man to cross the 0.15 Km train is = 6 s = (6/3600) h
Using, Distance = speed X time 0.15 = (v + 6) X (6/3600) v = 84 Km/h
So, the speed of the train is 84 Km/h
Q. The velocity of a car in 10 sec. changes from 10 m/s to 50 m/s. What will be its acceleration?
Ans: a = a =
Q. A train acquire velocity of 80km/h in just half an hour after start. Find its acceleration?

Ans: u = 0m/s ; v = = a = a =
Q.A Car is travelling at 20 m/s along a road. A child runs out into the road 50 m ahead and the car driver steps on the
brake pedal . What must the car's deceleration be if the car is to stop just before it reaches the child?
Ans: u = 20m/s ; s = 50 m v = 0 m/s Using, 2as = v2
- u2
a = = = - 4 m/s2
Acceleration = -4 m/s2
or, retardation = 4 m/s2
Q.A Car is travelling along the road at 8m/s. It accelerates at 1m/s2
for a distance of 18m. How fast is it travelling?
Ans: applying; v2
= u2
+ 2as v2
= 64 + 2x8 v2
= 100 So, V= 10m/s
Q. Stone dropped in a well hits water surface after 2 sec. What is the depth of well and with what speed will the
stone hit water surface?
Ans: a = -g = -9.8m/s2
; t = 2 sec; u = 0 m/s: Using s = u t + at2
you get s = - 19m or simply = 19 m
Using, you get v = 19.6m/s
Q. A ball hits a wall horizontally at 6m/s. it rebounds horizontally at 4.4m/s. The ball is in contact with the wall for
0.04 s. What is the acceleration of the ball?
Ans: u = 6m/s v= 4.4m/s t = 0.04sec. Using a = = 2
Q. A train starting from rest move with a uniform acceleration of 0.2m/s2
for 5 min calculate the speed acquired and
the distance travelled in this time?
Ans: u=0m/s a=0.2m/s2
t=5 min. = 300s
Using, s = s = u t + at2
s=0 + 1/2x 0.2x (300)2
=0+900=900m
Using, you get v = 60m/s
Q. Q. A car acquires a velocity of 180km/h in 20sec starting from rest find a) acceleration b) average velocity c)the
distance travelled in this time
Ans: a) At first we have to change speed in m/s, s = 180km/hr = 50m/s ; Given, u=0m/s , v = 50 m/s, t=20s
Using a = we get , a=50/20 = 2.5m/s2
(b) Average velocity = = 50 /2 => 25m/s

(c) Distance (s) = u t + at2
= 0 + 1/2 at2
= 1/2 x 2.5x(20)2
Weget, S = 500m
Q. 2 stones are thrown vertically upwards simultaneously with their initial velocities u1 and u2 respectively. Prove
that the heights reached by them would be in ratio (u1)2
: (u2) 2
Ans: Using, . At the top final velocity will be zero.
Using, or, h = but v = 0m/s and a = -g we get, h=
Therefore: h1 = and h2 = =

Ix physics motion_and_rest_solved_numerical

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