Numericals of Motion Class 9 – solved
Last updated on July 26th, 2023 at 01:05 pm
This page consists of interesting Numericals of Motion Class 9. These Numericals & questions are based on equations of motion (suvat equations). The solved answers to the Physics Numericals for class 9 motion are provided with step-by-step explanations.
Equations or formulas required to solve the motion numerical
This is the equation of the final velocity v.
v=u + at ……………….. (1)
Now see the equation of the displacement s.
s= ut + (1/2)at2 ……….(2)
Another equation of displacement without using acceleration a.
displacement = average velocity * time = [(u+v)/2]. t ………..(3)
This is another equation of the final velocity v. Please note that this equation of final velocity doesn’t have any mention of time t.
v2=u2 + 2as………………(4)
=> s = (v2-u2)/(2a) ……………. (5)
Average velocity = (u+v)/2 ……. (6) for uniform acceleration.
Again, Average velocity = displacement/time ……… (7) for uniform acceleration.
Numericals of Motion Class 9 | solved numerical problems in physics class 9 motion
1) A boat has an acceleration of 4 m/s2. What would the final velocity of the boat be after 10 seconds if the initial velocity of the speedboat is 2 m/s?
Solution:
a = 4 m/s2
u = 2 m/s
t = 10 s
final velocity = v
v = u + at = 2 + 4×10 = 42 m/s
2) A vehicle has an acceleration of 10 m/s2. What would the final velocity of the vehicle be after 20 seconds if the initial velocity of the vehicle is 40 m/s? What is the displacement of the vehicle during the 20 s time interval?
Solution:
a = 10 m/s2
u = 40 m/s
t = 20 s
final velocity = v
v = u + at = 40 + 10×20 = 240 m/s
final velocity = 240 m/s
Displacement during the 20 s time interval = s
s = ut + (1/2)at2
=> s = 40×20 + (½)x10x202
=> s= 800 + 2000 = 2800 m
Displacement during the 10 s time interval = 2800 m
3) A child on a flat board starts from rest and accelerates down a snow-covered hill at 1 m/s2. How long does it take the child to reach the bottom of the hill if it is 15.0 m away?
Solution: initial velocity u = 0
acceleration a = 1 m/s2
distance = s = 15 m
time taken t = ?
s= ut + (1/2)at2
As here, u = 0, so we can write:
s= (1/2)at2
=> t2 =(2 s /a ) = (2×15)/1 = 30
time t = 5.48 sec
4. A car starts from rest and accelerates at +2.50 m/s2 for a distance of 150.0 m. It then slows down with an acceleration of –1.50 m/s2 until the velocity is +10.0 m/s. Determine the total displacement of the car.
Solution:
First part:
displacement = s = 150 m
We need to find out the final velocity(v) of this accelerating part.
v2 = u2 + 2as
=> v2 = 0 + 2×2.5×150 = 750
=> v = 27.4 m/s
This velocity becomes the initial velocity of the second part of the journey.
Second part:
u = 27.4 m/s
a = -1.5 m/s2
v= 10 m/s
v2 = u2 + 2as
102 = 27.42 + 2.(-1.5).s
=> 100 = 750 – 3s
s= 650/3 = 216.7 m
Second part displacement = 216.7 m
Considering a straight-line journey without a change of direction,
total displacement = (150 + 216.7) m = 366.7 m
5. A car accelerates uniformly from a velocity of 21.8 m/s to a velocity of 27.6 m/s. The car travels 36.5 m during this acceleration.
a) What was the acceleration of the car?
b) Determine the time interval over which this acceleration occurred.
Solution:
a)
final velocity v = 27.6 m/s
initial velocity u = 21.8 m/s
distance traveled s=36.5 m
We will use this equation: v2 = u2 + 2as
a = ( v2 – u2) /(2s) = (27.62 – 21.82)/(2×36.5) = 3.92 m/s2
Acceleration = 3.92 m/s2
b) time interval over which this acceleration occurred = t
v = u + at
=> t = (v-u)/a = (27.6-21.8) / 3.92 = 1.48 seconds.
6) A bus is moving with the initial velocity of ‘u’ m/s. After applying the brake, its retardation is 0.5 m/s2 and it stopped after 12s. Find the initial velocity (u) and distance traveled by the bus after applying the brake.
Solution:
retardation a = –0.5 m/s2
final velocity v= 0
t = 12 s
u = ?
Using the formula: v=u + at
0 = u + (-0.5)(12)
=> 0 =u -6
=>u =6 m/s
Distance traveled:
s= ut + (1/2)at2
=> s = 6×12 + (1/2)(-0.5)122
=>s = 72 – (1/2)x72
=> s = 36 m
7 ) A car travels from rest with a constant acceleration ‘a’ for ‘t’ seconds. What is the average speed of the car for its journey if the car moves along a straight road?
Solution
The car starts from rest, so u = 0
The distance covered in time t:
s= (1/2)at2
Average speed =Total distance/Time taken
So, Average speed = (1/2)at2 / t =at/2
8) At a distance L= 400m away from the signal light, brakes are applied to a locomotive moving with a velocity, u = 54 km/h. Determine the time taken by the loco to stop. Also, determine the position of the rest of the locomotive relative to the signal light after 1 min of the application of the brakes if its acceleration a = – 0.3 m/s2
Solution
Since the locomotive moves with a constant deceleration after the application of brakes, say it comes to rest in t sec.
We know,
v = u + at
Here u = 54 km/h = 54 x 5/18 =15 m/s
v = 0 at time t
given a = -0.3 m/s2
From v = u+at we get t =(v-u)/a =(0-15)/(-0.3) = (-15)/(-0.3) =50 s
The time taken by the loco to stop = 50 s
To find the distance traveled we can use this equation s = (v2-u2)/(2a)
here v=0
so, the distance traveled s = -u2/(2a) = – 152/[2x(-0.3)]= 375 m
[note: As we have found out the time of travel we could use this equation as well to determine the distance: s= ut + (1/2)at2 ]
Thus in 1 minute after the application of brakes, the locomotive will be at a distance = 400 – 375 = 25 m from the signal light.
Summary of this post
We covered Equations or formulas required to solve the motion numerical problems and also solved a set of motion Numericals from class 9 physics.
