NEET Physics Motion in A Straight Line Questions Solved

(3)                    

Time to reach highest point after 3sec = $\frac{7-3}{2}=2sec.$

Time to reach highest point from ground = 3+2 = 5sec.

As upward and downward motion are symmetric [Time taken from A$\to$B is same as time taken from B$\to$C]

So, time to reach highest point is 5s. Same time is required to reach from B to ground.

For downward motion,

$$\begin{gathered} \mathrm{u}=0 \\ \mathrm{a}=\mathrm{g} \\ \mathrm{t}=5\mathrm{s} \\ \mathrm{h}=\frac{1}{2}{\mathrm{gt}}^2 \\ =\frac{1}{2}\times 10\times 5^2=125 \mathrm{m} \end{gathered}$$

$a=\frac{dv}{dt}=\frac{dv}{ds}\times \frac{ds}{dt}=v\times \frac{dv}{ds}$

(1)

$\mathrm{Motion} \mathrm{of} \mathrm{A} \mathrm{wrt} \mathrm{B}$

$$\begin{gathered} \mathrm{Relative} \mathrm{velocity} \mathrm{of} \mathrm{A} \mathrm{wrt} \mathrm{B} =\overrightarrow{ {\mathrm{V}}_{\mathrm{A}}} -\overrightarrow{ {\mathrm{V}}_{\mathrm{B}}} \\ =\mathrm{v}- (\mathrm{vcos}60° ) \\ = \mathrm{v}/2 \end{gathered}$$

 $\mathrm{Time} =\frac{\mathrm{a}}{{\displaystyle \frac{\mathrm{v}}{2}}}=\frac{2\mathrm{a}}{\mathrm{v}}$

(1)

$$\begin{gathered} t=\sqrt{x}+3 \\ x={\left(t-3\right)}^2 \\ At t=0 sec, x_1=9m \\ At t=6 sec, x_2=9m \\ Displacement=x_2-x_1=0 \end{gathered}$$

Distance travelled in t second,

$$\begin{gathered} Here, initial velocity u=0 \\ h=\frac{g{t}^2}{2} ...... \left(1\right) \\ Dis\tan ce travelled in n^{th} second = u +\frac{a}{2}(2n -1) \\ Dis\tan ce travelled in last second= 0 + \frac{g}{2}(2t-1) =\frac{9h}{25} ..............\left(2\right) \\ Equating h from equation 1 and 2 \\ \frac{25g(2t-1)}{18}=\frac{g{t}^2}{2} \\ 9t^2 -50t +25 =0 \\ Solving for t, \\ t =\frac{50 \pm \sqrt{{50}^2-4\times 9\times 25}}{2\times 9} = \frac{10}{18}s or 5 s \\ Time has to be greater than 1 second, so t=5s \\ Putting value of t in eqn. 1 \\ h=\frac{g{\left(5\right)}^2}{2} = 122.5 m \end{gathered}$$

(2)

Upward journey                                                                   Downward journey

Net acceleration = -(a+g)                                          Net acceleration = g - a
-ve indicates retardation

$$\begin{gathered} {\mathrm{t}}_{\mathrm{v}}= \sqrt{\frac{2\mathrm{h}}{\mathrm{g}+\mathrm{a}}} \\ {\mathrm{t}}_{\mathrm{D}}= \sqrt{\frac{2\mathrm{h}}{\mathrm{g}-\mathrm{a}}} \\ \frac{{\mathrm{t}}_{\mathrm{U}}}{{\mathrm{T}}_{\mathrm{D}}}= \sqrt{\frac{\mathrm{g}-\mathrm{a}}{\mathrm{g}+\mathrm{a}}}= \sqrt{\frac{8}{12}}= \sqrt{\frac{2}{3}} \end{gathered}$$