Motion-in-one-dimension-solution-16-to-41-1

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A car moving with a speed of 18 kmph comes to rest, when it moves
through a distance of 100m. calculate
(i) its uniform retardation,
(ii) time taken to come to rest,
(iii) its average speed.
Given u = 18 kmph
A stone is dropped from the top of a tower 96 m tall. At the same instant
another stone is thrown vertically upwards from the foot of the tower with a
velocity of 24 ms-1. Where and when do the two stones meet?
Let the two stones meet at a
height x above the ground level.
Therefore, for the motion AB
downwards
for the motion from C to B,
A body having an initial velocity of 0.9 ms-1 and uniform retardation,
passes through the starting point in 18 seconds. Calculate its retardation.
Given, u = 0.9 m/s2, t = 18s, a = ?
A stoone travels through a distance of 4.9 m in 0.1 s, and 4,998 m in the next
0.1 s. Calculate g at the place.
s1 = 4.9 m, s2 = 4.998 m,
A particle starting from rest and moving with uniform acceleration, covers
distances s1, s2 and s3 in three consecutive equal intervals of time. Calculate
the ratio s1: s2 : s3.
Let s1 be the distance covered in time t, starting from rest. Then,
A stone is dropped into a well and the sound of splash is heard after 3.91 s. If the depth
of the well is 67.6 m , find the velocity of sound.
Let the time taken for the stone to reach the water be t second.
A body moving with uniform acceleration, covers 11 m during the 4 th
second and 15 m during the 6th second. Calculate its initial velocity and
acceleration.
The initial velocity and acceleration is
A freely falling body covers half of the total distance during the last
second of its travel, find
(a) the time of fall and
(b) the height from which the body falls.
Let h be the height from which the body falls in a time t.
Then, (h/2) is the distance
travelled in (t -1) second.
A body starting from rest covers a distance of 70 m in the 4th
second. Find
(i) the distance travelled in 4s,
(ii) velocity at the end of 3s.
Given, u = 0, s3 = 70m, n = 4
Distance travelled in 4s is
(t = 4, u = 0, a = 20 m/s2)
Velocity at the end of 4s is
v = u + at
= 20 x 4 = 80 m/s.
A body is thrown up with a velocity of 78.4 ms-1. Find how high it will rise
and how much time it will take to return to its point of projection?
At the position of maximum
height s, v = 0
From the equation v2 = u2 +
2gs, we get,
Let t be time taken to reach the
maximum height.
Then,
A body starting from rest accelerates uniformly to attain a velocity of 20 ms-1 in
10 s. Compare the distance travelled by it in first 5 s and next 5s. What is the
distance travelled by it in the last second of its motion?
V = u + at
20 = 0 + a x 10 or a = 2
Let s1 and s2 be the distances
covered by it in the first and next 5
s respectively.
Velocity attained at the end of 5 s is
Determine the initial velocity and the acceleration of a particle travelling with uniform
acceleration in a straight line if it travels 55 m in the 8th sec and 85 m in the 13th sec. of
its motion.
A body is dropped from the top of a tower of height 100 m. Simulataneously
another body is thrown vertically upwards with a velocity of 25 ms-1. When
will the two bodies meet?
Let the bodies meet t seconds after the bodies begin their journeys.
If s1 and s2 be the distances
moved by the two bodies, then s1 +
s2 = 100
For the body dropped
For the other body
A bullet moving with a velocity of 80 m/s can just penetrate 5 wooden planks
of equal thickness. How many such planks can it penetrate if the initial
velocity is 124 m/s?
Let the thickness of each plank be x
Therefore, 8 = 5x
A stone thrown vertically up is back in the hand of the thrower at the end
of 4 seconds. If 'g' at the place is 9.8 m/s2, Calculate the velocity of the
throw and the greatest height reached by the stone.
Total time = 4s
The time taken to reach
maximum height = t = 2s
g = 9.8 m/s2
v=0
v = u + gt
The greatest height reached is
v = u + gt
0 = u - 9.8 x 2
u = 19.6 m/s
A stone is dropped from an aeroplane flying horizontally at 100 ms-1, when it
is vertically above a point A on the ground. If the stone reaches a point B on
the ground, calculate the distance AB. Height of the aeroplane is 78.4 meters
from the ground. (g = 9.8 ms-2)
O is the point of projection of the
stone and OA = 78.4 m. OB is the
actual path along which the
stone moves, until it reaches the
ground. But the time taken by
the stone to reach the earth is
equal to that taken by a body to
move along the vertical and
cover the path OA.
Given , u = 0, s = 78.4, g = 9.8, t = ?
The horizontal velocity of the
stone is uniform.
s = ut
= 100 x 4 = 400m
... AB = 400m.
A body is projected vertically upwards with a velocity of 20 ms-1. Calculate
(i) maximum height reached,
(ii) time taken to reach the maximum height,
(iii) velocity midway during the upward journey.
(i) v2 - u2 = -2gS
(ii) v = u - gt
(iii) u = 20, S = 10.2, v = ?
v2 - u2 = -2gS
v2 - 400 = -2 x 9.8 x 10.2
v2 = 400 - 50 = 350
v = 18.7 ms-1
A stone from rest, describes, 34.3 m in the last second of its fall. Calculate the height
from which the stone fell and the time of fall. (g = 9.8 ms-2)
Let n be the time of fall. Then
A wooden block is dropped from the top of a cliff 200 m high.
Simultaneously a bullet is fired from the foot of the cliff upwards with a
velocity of 200 m/s. Where and after what time will they meet?
Let x be the height at which the
wooden block and bullet meet
after 't' seconds.
For the block moving vertically
downwards
For the bullet moving vertically
upwards
From equation (1) and (2), we get,
A stone is projected vertically upwards from a point on the ground with a
velocity of 20 m/s. Two seconds later another stone is projected from the
same point with the same velocity. When and where do they meet? (g = 10
m/s2)
Diagram
Time taken by the first stone to
reach the maximum height is
Distance travelled by first
stone in 2s.
Let the two stones meet at a
distance x from the
ground after a time 't' second.
From the maximum height, the
first stone starts falling down.
(Now g = 10 m/s2, s = 20 - x, u =
0
For the stone projected
vertically upwards
From equation (1) + equation
(2), we get,
20 = 20t
t = 1s
Substituting in equation (1)
20 - x = 5 x1
x = 15 m.
A rocket is fired vertically from the ground with resultant vertical
acceleration of 10 m/s2. The fuel is finished in 60 seconds and it continues
to move up. What is the maximum height reached?
Initial velocity of the rocket, u
= , a = 10 m/s2
Let h be the height covered by
the rocket in t = 60s.
Let v be the final velocity at
the end of 60 s.
... v = u + at
= 0 + 10 x 60 = 600 m/s
At the end of 60 s, the rocket
moves with an initial velocity of
600 m/s under gravity.
... u = 600 m/s. a = - 9.8 m/s2, v = 0
Let h' be the height covered by the
rocket when it moves under
gravity.
Therefore, total height reached by
the rocket is h + h'
= 18000 + 18367
= 36367 = 36.367 km.
A stone is dropped from the top of a tower. Another stone B is thrown
downwards from the same point 2s later with an initial speed of 30 ms-1. If
both the stones A and B reach the ground together, find the height of the
tower.
Consider a body thrown upwards
from the top of the tower with a
velocity u, it reaches the ground in
't' seconds.
Net displacement = h = height of
the tower.
Initial velocity for this
displacement = - u
Acceleration a = g
Time taken = t second.
g = 10 ms-2
Let h be the height of the tower.
For the stone A,
For the stone B,
A stone is projected vertically up from the top of a tower at the velocity of
9.8 m/s. It reaches the foot of the building in 5 seconds. What is the height
of the tower and velocity of the stone when it reaches the ground? (g = 9.8
m/s2)
u = - 9.8 m/s, t = 5s, g = 9.7 m/s2
Net displacement = h = height of
the tower.

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