Explanation
Apply equation of kinematic
s=ut+12at2
t2+2uta−2sa=0
t2+pt−ks=0
Where, p=2uaandk=2a
t=−p±√p2+ks2
Hence, time is directly proportional to root of displacement.
Let assume initial velocity is u and constant acceleration a. using s=ut + (0.5)at2,
for first 5 second
10 = 5u + (0.5)a(25)
for next 3 seconds, t=8 and s=20
20=8u+(0.5)a(64)
Solving,
a=13 and u=76
Now for next 2 seconds, t=10 and S′:
S′=(10)76+(0.5)13(100)
S′ = 1706
Now distance travelled in 2 sec = 1706−20=506 meter
Given that,
Initial velocity u=20m/s
Final velocity v=0
Now, from equation of motion
v=u−gt
0=20−10t
t1=2010
t1=2s
Total time
t=2t1
t=2×2
t=4s
Hence, the time is 4 s
Given,
Initial velocity =u
After 2 cmpenetration, velocity v=u2
v2−u2=2as
(u2)2−u2=2a×2
a=−3u216
When, Final velocity v1=u4
Apply kinematic equation of motion
v21−u2=2as1
(u4)2−u2=2(−3u216)s1
s1=2.5 cm
Hence, for final velocity to be one fourth of initial velocity, body penetrate 0.5 cm more in wood.
Horizontal velocity, vx=20ms−1
Vertical initial velocity, uy=25ms−1
Acceleration of gravity, ay=−10ms−2
vy=uy+ayt
vy=25−10×2=5ms−1
Hence, velocity 20ms−1 horizontal and 5ms−1 upward
Mass m=2kg
Initial velocity u=0m/s
Final velocity v=20m/s
Time t=4s
Now, the acceleration is
v=u+at
a=v−ut
a=204
a=5m/s2
Now, the force is
F=ma
F=2×5
F=10N
Now, the displacement is
s=0+12×5×16
s=40m
Now, the work done is
W=F⋅s
W=10×40
W=400J
Now, the power is
P=Wt
P=4002
P=200watt
Hence, the power is 200 watt
Acceleration a=2m/s2
Retardation a=−3m/s2
Time t=10s
The maximum velocity attained be v at t
v=0+2×t
v=2tm/s.....(I)
Now, again it reaches velocity 0 after 10 s with retardation
0=v−3(10−t)
Now from equation (I)
2t−3(10−t)=0
2t−30+3t=0
5t=30
t=6s
Now, put the value of t in equation (I)
v=2t
v=2×6
v=12m/s
Hence, the maximum speed is 12 m/s
Final velocity v=0m/s
Acceleration a=−10m/s2
Now, the time by ball reach its maximum height
From equation of motion
t=2s
So, the time taken by the same ball to return to the hands of the juggler is
=2ug
=2×2010
=4s
So, he is throwing the balls after 1 sec each,
Let at some instant he throws ball number 4.
Now, Before 1 s of throwing it, he throws ball 3.
So, the height of ball 3 is
h3=20×1−12×10×(1)2
h3=15m
Before 2s, he throws ball 2, so the height of ball 2 is
h2=20×2−12×10×4
h2=20m
Before 3s, he throws ball 1, so the height of ball 1 is
h1=20×3−12×10×3×3
Hence, the height of balls in air from the ground will be 15 m, 20 m, 15 m and 0 m
Hint: Apply kinematical equations.
Correct Option: Option A
Explanation for correct answer:
Step 1: Find the length of train in terms of velocity and acceleration.
L=(2u)2+u22a=3u22a ------(1)
Step 2: Find the velocity of midpoint.
L2=v2−u22a
⇒3u24a=v2−u22a
⇒v=√2.5u m/s
So, the velocity with which, the middle point of the train passes the same point is √2.5u m/s.
Acceleration, a=dvdt=4−2t
At t = 0 s , a=4m/s2
At t = 1 s, a=2m/s2
At t = 2 s, a=0
At t = 3 s, a=4−6=−2m/s2
Hence, the acceleration of the body is initially positive but after some time it becomes negative.
Height h=5m
g=10m/s2
Now, for first drop
s1=ut+12gt2
5=0+5t2
t=1sec
It means that the third drop leaves after one second of the first drop. Or, each drop leaves after every 0.5 sec
Now, distance covered by the second drop in 0.5 s
s2=ut+12gt2
s=0+12×10×0.5×0.5
s=1.25m
Therefore, the distance of the second drop above the ground
s=s1−s2
s=5−1.25
s=3.75m
Hence, this is the required solution
a=btn
dvdt=btn
dv=btndt
v=btn+1n+1........(1)
drdt=btn+1n+1
dr=bn+1∫tn+1dt
r=bn+1tn+2n+2
r=Vtn+2(from1)
att=1s
V=(n+2)r
The formula for time of flight is
T=2usinθg
It is given that time is 2 seconds. So,
2=2usinθg
g=usinθ
Squaring both sides
g2=u2sin2θ........(1)
Height travelled is
H=u2sin2θ2g
H=g22g(from1)
H=g2=5m
Mass of ship, M=3\times {{10}^{7}}\,kg
Force, F=5\times {{10}^{4}}\,N
Distance, s=3\,m
Acceleration, a=\dfrac{F}{M}=\dfrac{5\times {{10}^{4}}}{3\times {{10}^{7}}}=\dfrac{1}{600\,}\,m{{s}^{-2}}
{{v}^{2}}-{{u}^{2}}=2as
v=\sqrt{2as}=\sqrt{2\times \dfrac{1}{600}\times 3}=0.1\,m{{s}^{-1}}
Hence, velocity of ship 0.1\,m{{s}^{-1}}
Distance s=45\,m
Initial velocity u=0
We know that,
By using equation of motion
{{v}^{2}}={{u}^{2}}+2as
{{v}^{2}}=0+2\times 9.8\times 45
{{v}^{2}}=882\,m/s
v=29.7\,m/s
Hence, the minimum speed is 29.7\ m/s
It is given that, the body reaches the maximum height in 6s. Let u be the initial velocity. At maximum height the velocity is zero. Using the equation of motion as
0=u-(10)(6)
u=60\,m/s
Distance travelled in nth second is given by:
{{S}_{n}}=u+\left( \dfrac{a}{2} \right)\left( 2n-1 \right)
Displacement in 1st second is {{S}_{1}}=60-\left( \dfrac{10}{2} \right)(2-1)=55\,m
Displacement in 7th second is {{S}_{2}}=60-\left( \dfrac{10}{2} \right)(14-1)=-5\,m
So, \dfrac{{{S}_{1}}}{{{S}_{2}}}=\dfrac{55}{5}=\dfrac{11}{1}
The ratio is 11:1.
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