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+12gt2
122.5=12×10×t2
⇒t=5sec (approx)
Time to be taken by the second ball to reach the ground =52=3sec.
If u be its initial velocity then,
122.5=u×3+12gt2=3u+12×10×9
3u=122.5−45=77.5
u=26 (approx)
u=0,v=144km/hour=144×518m/sec=40m/sec
v=u+at
⇒a=v−ut=40−020=2m/sec2
∴s=ut+12at2
=12×2×(20)2=400m
A bird flies for 4 s with a velocity of |t−2|m/s in a straightline,Where t is time in seconds.It covers a distance of
Max. Speed , v=u+at=0.2t,
v=0.2t1------------(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=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
Please disable the adBlock and continue. Thank you.