Explanation
Given,
Velocity v1 achieve in time t1,due to acceleration, a1=0.2ms−2
v1=u1+at1
⇒v1=0+0.2t1
Velocity reduce to zero in time t2, due to retardation, a2=−0.4ms−2
v2=v1+at2
⇒0=v1−0.4t2
Add time
t1+t2=v10.2+v10.4
⇒30×60=v1×7.5
⇒v1=240m/s
Apply kinematic equation
v2−u2=2as
s=v2−u22a
Total distance S1+S2=v122a1+v122a2=24022×0.2+24022×0.4=216km
A ball dropped from the top of the tower falls first half height of the tower in 10s. The total time spent by a ball in the air is [Takeg=10m/s2]
Let the height of tower is h. Now considering the equation as
s=ut+12at2
h=ut+12at2...........(1)
here,h=h2
In 10 sec, object is reaching half the height of tower.
Initially the body is rest so u = 0
It is also given that a = 10 ms-2
Time to reach first half height of tower t = 10 sec
Put the values in (1)
h2=0+12×10×102
h=1000m
The total time spent by a ball in the air is given by the relation as
1000=12×10×t2
t=10√2
t=14.14sec
A body moving along the positive x−axis with uniform acceleration of −4ms−2. Its velocity at x=0 is 10ms−1. The time taken by the body to reach a point at x=12cm is:
Let,
Constant acceleration, a
Initial velocity u
Apply kinematic Equation
v=u+at
a=v−ut
Compare acceleration at both event at time, t1&t2
v2−ut1=v3−ut2
⇒30−u2=60−u4
⇒u=0
Initial velocity u=zero
The acceleration when the mass is going up is given as,
v2=u2+2as
(20)2=(0)2+2as
a=200s (1)
The acceleration when the mass is going down is given as,
s=12at2
s=12a(10)2
a=s50 (2)
From equation (1) and (2), we get
200s=s50
s=100m
Substitute the value of s in the equation (2), we get
a=10050
=2m/s2
The weight is given as,
W=0.5×2
=1N
Thus, the weight of the body on that planet is 1N.
Force on crate, F=575N
Mass of crate, m=50kg
Coefficient of friction, μ=0.25
From the equilibrium of forces
ma=F−μmg
a=F−μmgm
a=575−0.25×50×9.8150
a=9.047m/s2
Acceleration of block is 9.047m/s2
Let distance covered by A be S1
Let distance covered by B be S2
B will catch A when S1 = S2
S = ut + 1/2 * a * t^2
Therefore,
40t + (1/2 * 0 * t^2) = 0t + 1/2 * 4 * t^2
40t = 2t^2
20t = t^2
Therefore, t = 20
Given that,
Mass m1=3kg
Mass m2=2kg
Now, the initial acceleration a of the system
a=3g−2g3+2
a=g5
At 5 sec
Now, from equation of motion
v=0+g5×5
v=9.8m/s
When the string breaks at t=5 s,
The mass 2 kg more under gravity
So, height is
h=v22g
h=(9.8)22×9.8
h=4.9m
Hence, the height is 4.9 m
Initial velocity, u=(2ˆj+ˆk)m/s
Force, F=(2ˆi+ˆj+3ˆk)N
→a=→Fm=2ˆi+ˆj+3ˆk1=(2ˆi+ˆj+3ˆk)m/s2
→v=→u+→at
→v=(2ˆj+ˆk)+(2ˆi+ˆj+3ˆk)×2
→v=(4ˆi+4ˆj+7ˆk)m/s
|→v|=√42+42+72=9m/s
Final velocity is 9m/s
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