Motion in Straight Line

If the man did not run, the bus would be at a distance s₁ at time t given by

   s₁ = 6 + $\frac{1}{2}\mathrm{\alpha }$ t²

       = 6 + $\frac{1}{2}$ × 3 × t²

   s₁ = 6 + $\frac{3}{2}$t²

If v is the speed of man, he would cover a distance s₂ = vt in time t.

To catch the bus,

           s₁ = s₂

     6 + $\frac{3}{2}$ t² = v t

⇒     t² $-$ $\frac{2 \mathrm{v}}{3}$t + 4 = 0

which gives

         t = $\frac{2 \mathrm{v}}{6} \pm {\left[ \frac{4 {\mathrm{v}}^2}{9} - 16 \right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$  

Now t will be real if  $\left(\frac{4 {\mathrm{v}}^2}{9} - 16\right)$ is positive or zero.

Minimum v corresponds to $\frac{4 {\mathrm{v}}^2}{16} - 16$= 0 which gives v = 6 m s^-1        

Two identical charged spheres suspended from a common point by two massless strings of length L.

In ,



The charge begins to leak from both the sphere at a constant rate. As a result, the spheres approach each other with velocity v.

Therefore, equation (i) can be rewritten as,

$$\begin{gathered} \mathrm{We} \mathrm{have}, \mathrm{distance} \mathrm{travelled} \mathrm{by} \mathrm{particle} 1 \mathrm{increases} \mathrm{with} \mathrm{time} \mathrm{as} \\ {\mathrm{x}}_1 = \frac{1}{2} {\mathrm{at}}^2 \\ \mathrm{Distance} \mathrm{travelled} \mathrm{by} \mathrm{particle} 2 \mathrm{is} \mathrm{proportional} \mathrm{to} \mathrm{t} \\ {\mathrm{x}}_2 = \mathrm{vt} \\ \therefore {\mathrm{x}}_1 - {\mathrm{x}}_2 = \frac{1}{2} {\mathrm{at}}^2 - \mathrm{vt} \\ \Rightarrow \mathrm{Graph} \mathrm{of} \left({\mathrm{x}}_1 - {\mathrm{x}}_2\right) \mathrm{vs} \mathrm{t} \mathrm{is} \mathrm{parabola}. \\ \mathrm{At} \mathrm{t} = 0 , {\mathrm{x}}_1 - {\mathrm{x}}_2 = 0 \\ \mathrm{Also} \mathrm{slope} = \frac{∆\mathrm{x}}{∆\mathrm{t}} = \mathrm{at} - \mathrm{v} \\ \mathrm{For} \mathrm{t} < \frac{\mathrm{v}}{\mathrm{a}} , \mathrm{the} \mathrm{slope} \mathrm{is} \mathrm{negative} \\ \mathrm{For} \mathrm{t} = \frac{\mathrm{v}}{\mathrm{a}} , \mathrm{the} \mathrm{slope} \mathrm{is} \mathrm{zero} \\ \mathrm{For} \mathrm{t} > \frac{\mathrm{v}}{\mathrm{a}} , \mathrm{the} \mathrm{slope} \mathrm{is} \mathrm{positive} \\ \mathrm{Hence}, \mathrm{option} \left(\mathrm{c}\right) \mathrm{is} \mathrm{correct}. \end{gathered}$$

x = 5t – 2t²    y = 10t



Acceleration of particle at t = 2 s is = –4 m/s²

Given:- 

        x = a e^-αt + be^βt

We know that

   velocity v = $\frac{\mathrm{dx}}{\mathrm{dt}}$

              v  =  $\frac{\mathrm{d} }{\mathrm{dt}}\left( {\mathrm{ae}}^{-\mathrm{\alpha t}} + {\mathrm{be}}^{\mathrm{\beta t}} \right)$

                  = -a$\mathrm{\alpha }$ e^-αt + bβ e^βt

              v = $-\mathrm{a} {\mathrm{\alpha e}}^{-\mathrm{\alpha t}} + \mathrm{b\beta } {\mathrm{e}}^{\mathrm{\beta t}}$

               v = A + B

where A = - a$\mathrm{\alpha } {\mathrm{e}}^{-\mathrm{\alpha t}}$ , B = bβ e^-βt

The value ot term A = -a$\mathrm{\alpha } {\mathrm{e}}^{-\mathrm{\alpha t}}$ increases and of term B = bβ e^-βt increases with time. As a result velocity goes on increasing with time. 

To move up with an acceleration 'a', the monkey will push the rope downward with a force

               F = ma = 30 a

Now,  T_max = mg + ma = 30 × 10 + 30 a

⇒        450 = 300 + 30 a

⇒            a = 5 m/s²

Hence, the rope will break if the monkey climbs up with an acceleration more than 5 m/s².

There are two vectors P and Q.

It is given that,


Let, angle between P and Q is ,

Given:- r = 25 cm

          r = 0.25  m

  ω = 13 rad/s

   Linear speed  = Radius × angular speed

            V = rω

          = 0.25 × 13

        3.25 m/s

Given:-

      A + B = C

Since, B = C $-$ A

Also

   $\left|\mathrm{C}\right| \perp \left|\mathrm{A}\right|$

    B² = 2 A²

⇒   B = $\sqrt{2}$ A

Now

    A² + B² + 2A B cosθ = C² = A²

∴          $\cos \mathrm{\theta } = - \frac{1}{\sqrt{2}}$

This gives

       $\mathrm{\theta } = \frac{3\mathrm{\pi }}{4}$ rad

Let balls meet after t s. The distance travelled by the ball coming down is 

          s₁ = $\frac{1}{2}$ gt²

Distance travelled by the other ball

        s₂ = 40 t $-$ $\frac{1}{2}$ gt²

         s₁ + s₂ = 100 m

∴        $\frac{1}{2}$ gt² + 40 t $-$ $\frac{1}{2}$gt² = 100m

                  t = $\frac{100}{40}$

∴                 t = 2.5 s