Physics Motion in Straight Line

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

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

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

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

At maximum height velocity (v) = 0

So, momentum = mv = m * 0 = 0

According to question, we can write

10 = $\frac{1}{2}a{t}^2...............\left(1\right)and$

$10+x=\frac{1}{2}a{\left(2t\right)}^2................\left(2\right)$

$\Rightarrow X=\frac{1}{2}A\left[{\left(2t\right)}^2-t^2\right]=3\left(\frac{1}{2}a{t}^2\right)=30m$

According to question, we can write

 10 = $\frac{1}{2}a{t}^2...............\left(1\right)and$                        

$10+x=\frac{1}{2}a{\left(2t\right)}^2................\left(2\right)$  

$\Rightarrow X=\frac{1}{2}A\left[{\left(2t\right)}^2-t^2\right]=3\left(\frac{1}{2}a{t}^2\right)=30m$  

Let S be the distance between Two ends a be the constant acceleration

As we know $v^2-u^2=2aS$

or, $aS=\frac{v^2-u^2}{2}$

Let $v_c$  be velocity at mid point.

Therefore, $v_c^2-u^2=2a\frac{S}{2}$

$v_c^2=u^2+aS$

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

$v_c=\sqrt[]{\frac{u^2+v^2}{2}}$

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,

Time taken by trains to collide = 300/150 = 2hrs
Speed of bird = 200 km/hr
Total distance travelled by bird = 200 * 2 = 400 km

For part AM, slope of v – t graph is constant but negative. For part MB, slope of v – t graph is constant but positive.