Explanation
Let t be the time in which the lizard will catch the insect.
Distance traveled by lizard = 21 + distance traveled by insect.
ut+12at2=21+vt
0+122t2=21+20t
t2−20t−21=0
solving this t=21,−1
t cannot be -ve.
t=21sec.
v2=2f(d+n)=2f′d
v=f′(t)=(m+t)f
eliminate d and m we get (f′−f)n=12ff′m2
Time taken by the first object to reach the ground = t, so
122.5 = ut + \displaystyle\frac{1}{2} g{t}^{2}
122.5 = \displaystyle\frac{1}{2} \times 10 \times {t}^{2}
\Rightarrow t = 5 sec (approx)
Time to be taken by the second ball to reach the ground = 5 2 = 3 sec.
If u be its initial velocity then,
122.5 = u \times 3 + \displaystyle\frac{1}{2} g{t}^{2} = 3u + \displaystyle\frac{1}{2} \times 10 \times 9
3u = 122.5 -45 = 77.5
u = 26 (approx)
u = 0, v = 144 {km}/{hour}= 144 \times \displaystyle\frac{5}{18} {m}/{sec} = 40 {m}/{sec}
v = u + at
\Rightarrow a = \displaystyle\frac{v-u}{t} = \displaystyle\frac{40-0}{20} = 2 {m}/{{sec}^{2}}
\therefore s = ut + \displaystyle\frac{1}{2} a{t}^{2}
= \displaystyle\frac{1}{2} \times 2 \times {\left(20\right)}^{2} = 400 m
A bird flies for 4 s with a velocity of \left| {t - 2} \right|m/s in a straightline,Where t is time in seconds.It covers a distance of
Max. Speed , v = u + at = 0.2 t ,
v = 0.2 t_1 ------------(1)
Dec --- v = u –at
0 = v – 1.0 t_2 , v = t_2 ----(2)
{t_1} = \dfrac{{{t_2}}}{{0.2}} = 5{t_2}\,\,\,\,
S = ut + \dfrac{1}{2} at^2
S = 2700 m 2700 = 0.1 [ (5t_2)^2 + (t_2)^2 - \frac{1}{2} {t_2}^2]
r= 80 cm
3 { t_2}^2 = 2700
t_2 = 30 sec, t_1 = 150 sec
Total time = t_1 + t_2 = 150 + 30 = 180 sec
Given,
Displacement of body, s=\dfrac{1}{2}a{{t}^{2}}
Kinematic equation, s=ut+\dfrac{1}{2}a{{t}^{2}}
On comparing, it is clear that acceleration is equal toa=g\,and initial velocity u=0.
Apply kinematic equation
v=u+at
v=0+gt
v=gt
Hence, velocity at time t is gt.
Given that,
Speed of bullet u=100\,m/s
Suppose that, the thickness of one plate is s
Now, from the equation of motion
{{v}^{2}}-{{u}^{2}}=-2as
0-{{u}^{2}}=-2as\left( s \right)
{{u}^{2}}\propto s
Now,
\dfrac{v_{2}^{2}}{v_{1}^{2}}=\dfrac{{{s}_{2}}}{{{s}_{1}}}
\dfrac{\left( {{\left( 2\times 100 \right)}^{2}} \right)}{{{\left( 100 \right)}^{2}}}=\dfrac{{{s}_{2}}}{{{s}_{1}}}
{{s}_{2}}=4{{s}_{1}}
{{s}_{2}}=4\times 2s
{{s}_{2}}=8s
Hence, the number of such planks is 8
v=ut+\dfrac{1}{2}\times gt^2
Differentiate y with respect to t which gives velocity as dy/dt = v
\dfrac{dv}{dt} = u + \dfrac{1}{2}g (2t)
v=u + gt -------------------Eqn (1)
Differentiate v with respect to t which gives velocity as d2y/dt2 = dv/dt = a
\dfrac{dv}{dt} = 0 + g
a = g ----------------------Eqn (2)We know that force acting on a mass m is given by F = ma
Now from Eqn (2), substitute a =g, Hence, F = mg
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