Physics Motion in Straight Line

v =? t +? t²
ds/dt =? t +? t²
? this =? (? t +? t²)dt from 1 to 2
s = [? t²/2 +? t³/3] from 1 to 2
s = [? (2)²/2 +? (2)³/3] - [? (1)²/2 +? (1)³/3]
s = [2? + 8? /3] - [? /2 +? /3]
s = 3? /2 + 7? /3

By t = mx² + nx
dt/dx = 2mx + n
V = dx/dt = 1/ (2mx + n)
a = v (dv/dx) = V (d/dx (1/ (2mx+n) = V [-2m/ (2mx+n)²] = -2mV³
So, Retardation will be (2mv³)

Displacement in 4? second
= S? - S?
= (1/2)g [2n - 1]
= (1/2)g [2*4 - 1]
= (1/2) * 9.8 * 7
= 34.3m
As this distance matches with data given in question for position of next drop. So drops are falling at the rate of 1 drop/second.

Body is dropped from height 75m with initial velocity (upward) 10m/s
So, s = ut + ½ at²
-75 = 10t - ½ gt²
5t² - 10t - 75 = 0
By solving t = 5 sec.
In 5sec. balloon covered
h = 10 * 5 = 50m
Now height of balloon = 75 + 50 = 125m

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

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

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

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

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,

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

When second stone is released

Equation for first ball:

$x+20=10t+\frac{1}{2}g{t}^2$

Equation for second ball :

$x=\frac{1}{2}g{t}^2$

Using these two equation

20 = 10t t = 2 sec

$x=\frac{1}{2}*10*4=20m$

H = 20 + 20 + 5 = 45 m.

Initially, the body has zero velocity and zero slope. Hence the acceleration would be zero initially.

After that, the slope of v-t curve is constant and positive.

After some time, velocity becomes constant and acceleration is zero.

After that, the slope of v-t curve is constant and negative.

Air resistance resists the motion of an object. In this case, the net acceleration is lesser than 'g' and it shrinks as the speed increases. This makes the object to speed up more slowly. Ultimately, it reaches a constant terminal velocity which is lower for large-area ones and higher for heavy and streamlined ones.