Equation of motion
Distance and displacement
Distance: For a moving body, length of actual path followed. It is scalar quantity.
Displacement: shortest distance from initial position to final position with direction. It is a vector quantity.
Distance travelled ≥ displacement covered.
Note: Total displacement covered during certain time may be zero but distance travelled is never zero.
Distance and displacement
Distance: For a moving body, length of actual path followed. It is scalar quantity.
Displacement: shortest distance from initial position to final position with direction. It is a vector quantity.
Distance travelled ≥ displacement covered.
Note: Total displacement covered during certain time may be zero but distance travelled is never zero.
Note: A body is moving in circular path of radius R and it displaces certain angle θ in time t as in fig
OA + AB = OB
AB = OB - OA
ǀABǀ = ǀOB - OAǀ
= √(OB2 - 2OA OB cos θ + OA2)
= √(2R2 - 2R2 cos θ)
= √2R √(1 - cos θ)
= √2R √(1 - 1 + 2 sin2 θ/2)
Displacement, AB = 2R sin θ/2 = 2R sin wt/2
& Distance = R θ = wtR , where θ is in radian.
Q. A vehicle is moving along a circular track of radius 200 m at constant angular speed π / 30 rad/sec. The displacement covered by vehicle in 10 s will be
(a) 100 π / 3 (b) 100 (c) 10 π / 3 m (d) 100 π
Ans: (b)
Hint: Displacement = 2R sin θ/2
Speed
Rate of change of distance. It is a scalar quantity.
Avg. speed = (total distance) / (total time) = Δs / Δt
velocity
Rate of change of displacement. It is a vector quantity.
Avg. velocity = (total displacement) / (total time) = Δs / Δt
Acceleration
Rate of change of velocity. It is a vector quantity.
Avg. acceleration (a) = Δv / Δt
OA + AB = OB
AB = OB - OA
ǀABǀ = ǀOB - OAǀ
= √(OB2 - 2OA OB cos θ + OA2)
= √(2R2 - 2R2 cos θ)
= √2R √(1 - cos θ)
= √2R √(1 - 1 + 2 sin2 θ/2)
Displacement, AB = 2R sin θ/2 = 2R sin wt/2
& Distance = R θ = wtR , where θ is in radian.
Q. A vehicle is moving along a circular track of radius 200 m at constant angular speed π / 30 rad/sec. The displacement covered by vehicle in 10 s will be
(a) 100 π / 3 (b) 100 (c) 10 π / 3 m (d) 100 π
Ans: (b)
Hint: Displacement = 2R sin θ/2
Speed
Rate of change of distance. It is a scalar quantity.
Avg. speed = (total distance) / (total time) = Δs / Δt
velocity
Rate of change of displacement. It is a vector quantity.
Avg. velocity = (total displacement) / (total time) = Δs / Δt
Acceleration
Rate of change of velocity. It is a vector quantity.
Avg. acceleration (a) = Δv / Δt
Q. An object travels displacement s with 20 m/s and returns with velocity 30 m/s. Find avg. velocity and speed.
Soln:
Avg. velocity = (total displacement) / (total time) = 0/t = 0
total distance = 2s
total time (T) = s/20 + s/30
So Avg. speed = {total distance (2s)} / {total time(T)} = 2s / {s/20 + s/30} = 24 m/s
Soln:
Avg. velocity = (total displacement) / (total time) = 0/t = 0
total distance = 2s
total time (T) = s/20 + s/30
So Avg. speed = {total distance (2s)} / {total time(T)} = 2s / {s/20 + s/30} = 24 m/s
Q. Choose the current option.
(a) speed =ǀ velocity ǀ (b) speed > ǀ velocity ǀ (c) speed < ǀ velocity ǀ (d) speed ≥ ǀ velocity ǀ
Ans: (d)
Q. A person travels at constant velocity V1 due east and then he travels the same distance at constant velocity V2 due north. The avg. velocity will be
(a) √{(v12 + v22)/2} (b) ) 1/2 √(v12 + v22) (c) ) {√2(v1 v2) /(v1 + v2)} (d ) (2v1 v2) /(v1 + v2)
Ans: (c)
T = s/v1 + s/v2 = s(v1 + v2) / (v1 v2)
Vavg = (total displacement) / (total time)
= √2 s / {s (v1 + v2)/ (v1 v2)}
Vavg = {√2(v1 v2) /(v1 + v2)}
Q. Find out the speed of a particle if position of a particle varies like x = at2 and y = 3bt3
soln:
Vx = dx / dx = 2at i
Vy= dy / dt = 9 bt2 j
ǀVǀ = √ {(2at)2 + (9 bt2)2} & tan θ = Vy / Vx
= √ {(4a2t2 + 18 b2t4}
Motion under gravity
For upward motion For downward motion
v = u - gt v = u + gt
v2 = u 2 - 2gh v2 = u 2 + 2gh
h = ut - 1/2 gt2 h = ut + 1/2 gt2
ht = u - g/2 (2t - 1) ht = u + g/2 (2t - 1)
The value of g = 9.81 m/s2 = 981 cm/s2 = 32 ft/s2
Distance travelled by a body in a st. line in tth second is
St = u + a/2 (2t - 1)
Q. A body is thrown vertically upward from ground. While moving upward, the acceleration of the body without neglecting air resistance is
(a) equal to g (b) more than g (c) less than g (d) zero
Ans: (b)
Hint: F = mg + 6πȠrv
=> a = g + (6πȠrv)/m
=> a > g , when moves upward
(a) speed =ǀ velocity ǀ (b) speed > ǀ velocity ǀ (c) speed < ǀ velocity ǀ (d) speed ≥ ǀ velocity ǀ
Ans: (d)
Q. A person travels at constant velocity V1 due east and then he travels the same distance at constant velocity V2 due north. The avg. velocity will be
(a) √{(v12 + v22)/2} (b) ) 1/2 √(v12 + v22) (c) ) {√2(v1 v2) /(v1 + v2)} (d ) (2v1 v2) /(v1 + v2)
Ans: (c)
T = s/v1 + s/v2 = s(v1 + v2) / (v1 v2)
Vavg = (total displacement) / (total time)
= √2 s / {s (v1 + v2)/ (v1 v2)}
Vavg = {√2(v1 v2) /(v1 + v2)}
Q. Find out the speed of a particle if position of a particle varies like x = at2 and y = 3bt3
soln:
Vx = dx / dx = 2at i
Vy= dy / dt = 9 bt2 j
ǀVǀ = √ {(2at)2 + (9 bt2)2} & tan θ = Vy / Vx
= √ {(4a2t2 + 18 b2t4}
Motion under gravity
For upward motion For downward motion
v = u - gt v = u + gt
v2 = u 2 - 2gh v2 = u 2 + 2gh
h = ut - 1/2 gt2 h = ut + 1/2 gt2
ht = u - g/2 (2t - 1) ht = u + g/2 (2t - 1)
The value of g = 9.81 m/s2 = 981 cm/s2 = 32 ft/s2
Distance travelled by a body in a st. line in tth second is
St = u + a/2 (2t - 1)
Q. A body is thrown vertically upward from ground. While moving upward, the acceleration of the body without neglecting air resistance is
(a) equal to g (b) more than g (c) less than g (d) zero
Ans: (b)
Hint: F = mg + 6πȠrv
=> a = g + (6πȠrv)/m
=> a > g , when moves upward
F = mg - 6πȠrv
=> a = g - (6πȠrv)/m
=> a <g , when moves downward
When a body is released or dropped from the certain height h then,
=>Time taken, t = √(2h/g)
=>Velocity at height h' above ground, v = √{2g(h - h')}
=> Final velocity, v =√(2gh)
=> Height of the particle at any time h = H- g/2 (T - t)2
=> a = g - (6πȠrv)/m
=> a <g , when moves downward
When a body is released or dropped from the certain height h then,
=>Time taken, t = √(2h/g)
=>Velocity at height h' above ground, v = √{2g(h - h')}
=> Final velocity, v =√(2gh)
=> Height of the particle at any time h = H- g/2 (T - t)2
A body is dropped from a top of tower and falls freely
(i) distance covered by it after n sec = 1/2 g n2 i.e. α n2
(ii) distance covered in nth sec = 0 + 1/2 g (2n - 1) i.e. α (2n - 1)
(iii) velocity of body after n sec = 0 + gn i.e. α n
(i) distance covered by it after n sec = 1/2 g n2 i.e. α n2
(ii) distance covered in nth sec = 0 + 1/2 g (2n - 1) i.e. α (2n - 1)
(iii) velocity of body after n sec = 0 + gn i.e. α n
Q. A body is released from a height 45m. The distance covered by it in last 0.6 sec is
(a) 8.6 m (b)7.2 m (c) 16.2 m (d) 18.2 m
Ans: (c)
Hint: t = √(2h/g) =√(2 * 45 /10) = 3 sec.
h = ut + 1/2 gt2 = 0 + 1/2 * 10 * 2.42 = 28.8 m
distance covered in last 0.6 sec = (45 - 28.8) = 16.2 m
Q. A body is thrown vertically upward from earth surface with velocity 'u', which can attain a maximum height 'H'. What is its value at 3H/4 above earth?
(a) u/4 (b) 3/4 u (c) u/2 (d) √3/2 u
Ans: (c)
Hint: u' = √{2g(h - h') or put h' = 3H/4
v' = √{2g(H - 3H/4)} = = √{2gH/4} = √(2gH)/2 = u/2
Relative Velocity
It is a velocity of one body w.r.t. another body when both the bodies are in motion. Relative velocity of a w.r.t. B is
VAB = VA - VB
VAB = √{VA2 + VB2 + 2VAVB cos(180 - θ)}
Cases:
(1) when A & B are moving in same direction.
θ = 0o or cos 0 = 1
VAB = VA - VB
(2) when A & B are moving in opposite direction.
It θ = 180o
VAB = VA + VB
Q. A man walks in rain with a velocity of 5 km/h. The rain drops strike at him at an angle of 45o with the horizontal. The downward velocity of the raindrops will be
(a) 5 km/h (b) 4 km/h (c) 3 km/h (d) 1 km/h
Ans: (a)
(a) 8.6 m (b)7.2 m (c) 16.2 m (d) 18.2 m
Ans: (c)
Hint: t = √(2h/g) =√(2 * 45 /10) = 3 sec.
h = ut + 1/2 gt2 = 0 + 1/2 * 10 * 2.42 = 28.8 m
distance covered in last 0.6 sec = (45 - 28.8) = 16.2 m
Q. A body is thrown vertically upward from earth surface with velocity 'u', which can attain a maximum height 'H'. What is its value at 3H/4 above earth?
(a) u/4 (b) 3/4 u (c) u/2 (d) √3/2 u
Ans: (c)
Hint: u' = √{2g(h - h') or put h' = 3H/4
v' = √{2g(H - 3H/4)} = = √{2gH/4} = √(2gH)/2 = u/2
Relative Velocity
It is a velocity of one body w.r.t. another body when both the bodies are in motion. Relative velocity of a w.r.t. B is
VAB = VA - VB
VAB = √{VA2 + VB2 + 2VAVB cos(180 - θ)}
Cases:
(1) when A & B are moving in same direction.
θ = 0o or cos 0 = 1
VAB = VA - VB
(2) when A & B are moving in opposite direction.
It θ = 180o
VAB = VA + VB
Q. A man walks in rain with a velocity of 5 km/h. The rain drops strike at him at an angle of 45o with the horizontal. The downward velocity of the raindrops will be
(a) 5 km/h (b) 4 km/h (c) 3 km/h (d) 1 km/h
Ans: (a)
Q. A river is flowing from west to east at a speed of 5 m/min. In what direction should a man on south bank of the river capable of swimming at 10 m/ min in still water should swim to cross the river
(i) in shortest time (ii) along shortest path
Q. A person who can swim in still water at 5 km/h crosses a river, 1 km wide flowing at 3 km/h in least possible time. The time taken to cross the river is
(a) 20 min (b) 12 min (c) 15 min (d) 20 min
Ans: (b)
t = d / (v cos θ)
tmin = d / v = 1/5 hr = 12 min
Q. A person who can swim in still water at 5 km/h crosses a river, 1 km wide flowing at 3 km/h along shortest route. The time taken to cross the river will be
a) 10 min (b) 12 min (c) 15 min (d) 20 min
Ans: (c)
Q. A person has to start swimming at right angle to the stream of river flowing at velocity u in order to cross that river in least time.
Soln: vel. across water = √(v2 + u2)
least time taken = AB/v = BC / u = AC /√( v2 + u2)
Q. A stone is dropped from a balloon going up with a uniform velocity of 5 m/s. If the balloon was 50 m high when the stone was dropped. Find the time when the stone hits the ground. (g = 10 m/s2).
(a) 2.7 s (b) 3.7 s (c) -2.7 s (d) 4.7 s
Ans: (b)
Soln: Taking vertically upwards as +ve X-axis. (↑+ve ↓-ve)
x = ut + 1/2 at2
-50 = 5 × t + 1/2 × (-10) × t2
or 5 t2 - 5t - 50 = 0
or t2 - t - 10 = 0
t = [+ 1 ± √{1 - 4×1× (-10)} ] / 2 = (1 ± √41) / 2
=> t = 3.75, & t = - 2.7 s
Q. A heavy truck and a car are in motion and both have same k.E. Brakes are applied to produce equal retarding force. Which will travel a smaller distance before coming to rest.
(a) car (b) truck (c) both travel through same distance (d) None
Ans: (c)
(i) in shortest time (ii) along shortest path
Q. A person who can swim in still water at 5 km/h crosses a river, 1 km wide flowing at 3 km/h in least possible time. The time taken to cross the river is
(a) 20 min (b) 12 min (c) 15 min (d) 20 min
Ans: (b)
t = d / (v cos θ)
tmin = d / v = 1/5 hr = 12 min
Q. A person who can swim in still water at 5 km/h crosses a river, 1 km wide flowing at 3 km/h along shortest route. The time taken to cross the river will be
a) 10 min (b) 12 min (c) 15 min (d) 20 min
Ans: (c)
Q. A person has to start swimming at right angle to the stream of river flowing at velocity u in order to cross that river in least time.
Soln: vel. across water = √(v2 + u2)
least time taken = AB/v = BC / u = AC /√( v2 + u2)
Q. A stone is dropped from a balloon going up with a uniform velocity of 5 m/s. If the balloon was 50 m high when the stone was dropped. Find the time when the stone hits the ground. (g = 10 m/s2).
(a) 2.7 s (b) 3.7 s (c) -2.7 s (d) 4.7 s
Ans: (b)
Soln: Taking vertically upwards as +ve X-axis. (↑+ve ↓-ve)
x = ut + 1/2 at2
-50 = 5 × t + 1/2 × (-10) × t2
or 5 t2 - 5t - 50 = 0
or t2 - t - 10 = 0
t = [+ 1 ± √{1 - 4×1× (-10)} ] / 2 = (1 ± √41) / 2
=> t = 3.75, & t = - 2.7 s
Q. A heavy truck and a car are in motion and both have same k.E. Brakes are applied to produce equal retarding force. Which will travel a smaller distance before coming to rest.
(a) car (b) truck (c) both travel through same distance (d) None
Ans: (c)
Note: The stopping distance of a car is x when its speed is v, what will be the stopping distance when its speed is increased to nv during same time.
V2 = u2 + 2ax => a = v2 / 2x & x = v2 / 2a
when V = nv
then x' = v2 / 2a (n2 v2 ) / {2( v2/2x)} = n2 x
Q. The shortest distance over which a car can be brought to rest by applying breaks is 1m. If its speed is made 3 times, the car can be brought to rest over a distance of
(a) 9 m (b) 6 m (c) 3 m (d) 1 m
Ans: (a)
Hint: x' α n2 x
Note: A body is dropped from the top of a tower of height 'h' m. if simultaneously another ball thrown upward with velocity 'v'. After what time do they cross each other?
=> h = h1 + h2
or h = (0 + 1/2 gt2) + (vt - 1/2 gt2)
t = h/v
h2 = h - g/2 h2 / v2
Q. A ball is dropped from 100 m height. If simultaneously another ball thrown upward with velocity 50 m/s. When and where would they meet each other.
(a) 1 sec, 20 m (b) 2 s, 80 m (c) 3 s, 80 m (d) 4 s, 80 m
Ans: (b )
Hint: h/v = 100/50 = 2
Q. A passenger is x m away from the bus moving with a constant acceleration a. What is the minimum velocity of the passenger so that he can catch the bus?
Soln: vt = x + 1/2 at2
=> t = {v ± √(v2 - 2ax)} / a
t is real so v2 - 2ax ≥ 0
So, Vmin = √(2ax)
V2 = u2 + 2ax => a = v2 / 2x & x = v2 / 2a
when V = nv
then x' = v2 / 2a (n2 v2 ) / {2( v2/2x)} = n2 x
Q. The shortest distance over which a car can be brought to rest by applying breaks is 1m. If its speed is made 3 times, the car can be brought to rest over a distance of
(a) 9 m (b) 6 m (c) 3 m (d) 1 m
Ans: (a)
Hint: x' α n2 x
Note: A body is dropped from the top of a tower of height 'h' m. if simultaneously another ball thrown upward with velocity 'v'. After what time do they cross each other?
=> h = h1 + h2
or h = (0 + 1/2 gt2) + (vt - 1/2 gt2)
t = h/v
h2 = h - g/2 h2 / v2
Q. A ball is dropped from 100 m height. If simultaneously another ball thrown upward with velocity 50 m/s. When and where would they meet each other.
(a) 1 sec, 20 m (b) 2 s, 80 m (c) 3 s, 80 m (d) 4 s, 80 m
Ans: (b )
Hint: h/v = 100/50 = 2
Q. A passenger is x m away from the bus moving with a constant acceleration a. What is the minimum velocity of the passenger so that he can catch the bus?
Soln: vt = x + 1/2 at2
=> t = {v ± √(v2 - 2ax)} / a
t is real so v2 - 2ax ≥ 0
So, Vmin = √(2ax)
Q. A ball is thrown upwards with speed u from a top of a tower and it reaches the ground after time t1. If same ball is thrown downwards from the same point with same speed u, it reaches to ground after time t2. What time does it take to reach to ground dropped from same point.
Soln:
h = ut1 + 1/2 gt12 = ut2 + 1/2 gt22
or 1/2 g (t12 - t22 ) = u (t2 - t1 )
=> u = g/2 (t1 - t2 )
So, h = g/2 (t1 - t2 )t2 + 1/2 gt2
= g/2 t1 t2 - g/2 t22 + 1/2 gt2
. : h = g/2 t1 t2
Soln:
h = ut1 + 1/2 gt12 = ut2 + 1/2 gt22
or 1/2 g (t12 - t22 ) = u (t2 - t1 )
=> u = g/2 (t1 - t2 )
So, h = g/2 (t1 - t2 )t2 + 1/2 gt2
= g/2 t1 t2 - g/2 t22 + 1/2 gt2
. : h = g/2 t1 t2
again , h = 0 + d/2 gt2
so, g/2 t1 t2 = 1/2 gt2
=> t = √ t1t2
Q. A body when thrown upward with a certain velocity reaches to ground after 8s. If the same body is thrown downward from the same point with same velocity, it reaches to ground in 2 s. How much time does it take to reach to ground when it is dropped from the same point?
(a) 2 s (b) 4 s (c) 12 s (d) 10 s
Ans: (b )
Hint: t = √ t1t2 = √(8 × 2) = 4
so, g/2 t1 t2 = 1/2 gt2
=> t = √ t1t2
Q. A body when thrown upward with a certain velocity reaches to ground after 8s. If the same body is thrown downward from the same point with same velocity, it reaches to ground in 2 s. How much time does it take to reach to ground when it is dropped from the same point?
(a) 2 s (b) 4 s (c) 12 s (d) 10 s
Ans: (b )
Hint: t = √ t1t2 = √(8 × 2) = 4
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