Physics Motion in Straight Line

S₁ = $\frac{1}{2}$ a (p – 1)²

S₂ = $\frac{1}{2}$ ap²

S₁ + S₂ = $\frac{1}{2}$ a [ (p – 1)² + p²] = $\frac{1}{2}$ at²

$t=\sqrt[]{2p^2+1-2p}$

Dv = a (t + 1 – t)

a = 50 m/s²

$s=u+\frac{a}{2}\left(2\left(t+1\right)-1\right)$  

$120=u+\frac{50}{2}\left(2t+1\right)$  

u = 120 – 25 (2t +1)

In (t + 2)^th second

$s\text{'}=u+\frac{a}{2}\left(2\left(t+2\right)-1\right)$

= u + 25 (2t + 3)

= 120 – 25 (2t) – 25 + 25 (2t) + 75

S' = 170 m

Distance travelled by the body = Sum of areas under the curve (both below and above x-axis)

$\frac{1}{2}*3*10*2$

$=30m$

$V=\sqrt[]{180-16x}m/ sec$

$ac{c}^n=V\frac{dv}{dx}$

$={\left(180-16x\right)}^{1/2}\frac{d{\left(180-16x\right)}^{1/2}}{dx}$

= ${\left(18-16x\right)}^{1/2}*\frac{1}{2}{\left(180-16x\right)}^{1/2}*-16$

= $?8 m/se{c}^2$

Distance travelled by the body = Sum of areas under the curve (both below and above x-axis)

$\frac{1}{2}*3*10*2$

$=30m$

  $tan\alpha =\frac{v_{max}}{t_1}=a_1$

$tan\beta =\frac{v_{max}}{t_2}=a_2$

$\Rightarrow \frac{a_1}{a_2}=\frac{t_2}{t_1}\Rightarrow \frac{t_1}{t_2}=\frac{a_2}{a_1}$

Assume the length of train be I and its acceleration be a.

$v^2=u^2+2al\Rightarrow al=\frac{v^2-u^2}{2}$

Velocity when middle point crosses the post,

$V_m=\sqrt[]{u^2+2a\frac{\mathcal{l}}{2}}.=\sqrt[]{u^2+\frac{v^2-u^2}{2}}=\sqrt[]{\frac{u^2+v^2}{2}}$

Let 'h' be the height at which velocity becomes equal to magnitude of Acceleration

v = g = 10

v = u + at

10 = 0 + 10t

t = 1 sec

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

$=0*1+\frac{1}{2}*10*1*1$

= 5m