CBSE Questions for Class 11 Engineering Physics Motion In A Straight Line Quiz 1 - MCQExams.com

Let t be the time in which the lizard will catch the insect.

Distance traveled by lizard $$=$$ 21 + distance traveled by insect.

$$ut+\frac { 1 }{ 2 } a{ t }^{ 2 }=21+vt$$

$$0+\frac { 1 }{ 2 } 2{ t }^{ 2 }=21+20t$$

$${ t }^{ 2 }-20t-21=0$$

solving this $$t= 21 , -1 $$

t cannot be -ve.

$$t=21 sec.$$

$$\mathrm{v}^{2}=2\mathrm{f}(\mathrm{d}+\mathrm{n})=2\mathrm{f}'\mathrm{d}$$

$$\mathrm{v}=\mathrm{f}'(\mathrm{t})=(\mathrm{m}+\mathrm{t})\mathrm{f}$$

eliminate $$\mathrm{d}$$ and $$\mathrm{m}$$ we get $$(f^{'}-f)n=\frac{1}{2}ff^{'}m^{2}$$ 

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)

• $$v=u+at$$
• $$s=ut+\frac { 1 }{ 2 } at^{ 2 }$$
• $${ v }^{ 2 }{ =u }^{ 2 }+2as$$

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 to$$a=g\,$$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