Let, u cm/sec be initial velocity of bullet.
By question, velocity of bullet after pertaining 2.5 cm into target will be u/2 cm/sec.
Now if uniform retardation to penetration be f cm/sec²,
Then using formula, v² = u² ? 2fs, we get,
(u/2)² = u² ? 2f(2.5)
Or f = 3u²/20
Now let us assume that bullet can penetrate x cm into target.
Then final velocity of bullet will be zero after penetrating x cm into target.
Hence, using formula v² = u² ? 2fs we get,
0 = u² ? 2(3u²/20)(x)
Or 10 ? 3x = 0
Or x = 10/3 = 3.33(approx).
Therefore, required furr penetration into target will be
3.33 ? 2.5 = 0.83 cm.

We have, x = a + bt + ct²  ???.(1)
Let, v and f be velocity and acceleration of a particle at time t seconds.
Then, v = dx/dt = d(a + bt + ct²)/dt = b + ct   ???.(2)
And f = dv/dt = d(b + ct)/dt = c   ???.(3)
Clearly, when c > 0,implies f > 0.
Hence in this case particle moves withan uniform positive acceleration.

Let particle projected from A with velocity u = 196 m/sec and P be its position at time t where, AP = x m, if v m/sec be its velocity at P,n equation of motion of particle is,
dv/dt = -g [where, v = dx/dt] ???.(1)
Since v = u, when t = 0, hence from (1) we get,
_u∫^vdv = -g ₀∫^tdt
Or v ? u = -gt
Or dx/dt = u ? gt ???.(2)
Since x = 0, when t = 0, hence from (2) we get,
₀∫^x dx = ₀∫^x (u ? gt)dt
Or x = ut ? (1/2)gt² ???.(3)
Again, (1) can be written as,
dv/dx*dx/dt = -g
Or v(dv/dx) = -g ???.(4)
Since v = u, when x = 0, hence, from (4) we get,
_u∫^vvdv = -g ₀∫^xdx
Or v² = u² ? 2gx ???.(5)
If H m be greatest height of particle, n v = 0, when x = H; hence, from (5) we get,
0 = u² ? 2gH
Or H = u²/2g
= (196*196)/(2*9.8)   [as, g = 9.8 m/sec²]
= 1960 m.

f(x) = 10 ? 9x + 6x² ? x³
Thus, f?(x) = ? 9 + 12x ?3 x²
= -3(x² ? 4x + 3)
Or f?(x) = -3(x ? 1)(x ? 3) ???.(1)
If x > 3, n x ? 1 > 0 and x ? 3 > 0
Hence, (x ? 1)(x ? 3) > 0
Thus, from (1) it readily follows that, f?(x) < 0, when x > 3
So, f(x) decreases for values of x > 3.

Let initial velocity of particle be u cm/second and its uniform retardation be f cm/sec².
Furr assume that particle was in motion for t seconds.
By question, particle comes to rest after t seconds.
Therefore, using formula, v = u ? ft, we get,
0 = u ? ft
Or u = ft
Again, particle described 7cm in 5th second. Therefore, using formula
s_t = ut + 1/2(f)(2t ? 1) we get,
7 = u ? ½(f)(2.5 ? 1)
Or u ? 9f/2 = 7
Again, distance described in last second (i.e., in t th second) of its motion
= 1/64 * (distance described by particle in t seconds)
ut -1/2(f)(2t ? 1) = 1/64(ut ? ½(ft²))
Putting u = ft, we get,
f/2 = 1/64((ft²)/2)
Or t² = 64
Or t = 8 seconds.

We have, x = a + bt + ct²  ???.(1)
Now from (1) we get x = a, when t = 0.
This means that distance of particle from O at time t = 0 is a ft.
Thus, a represents initial position of particle.

Equation of given hyperbola is, 3x² ? 4y² = 12 ???.(1)
Differentiating both sides of (1) with respect to y we get,
3*2x(dy/dx) ? 4*(2y) = 0
Or dx/dy = 4y/3x
Therefore, equation of normal to hyperbola (1) at point (x₁, y₁) on it is,
y ? y₁ = -[dx/dy]_{(x1, y1)} (x ? x₁) = -4y₁/3x₁(x ? x₁)
Or 3x₁y + 4y₁x – 7x₁y₁ = 0
Now, if possible, let us assume that straight line
x + y + 7 = 0 ???..(2)
This line is normal to hyperbola (1) at point (x₁, y₁). Then, equation (2) and (3) must be identical. Hence, we have,
3x₁/1 = 4y₁/1 = -7x₁y₁/7
So, x₁ = -4 and y₁ = -3
Now, 3x₁² ? 4y₁² = 3(-4)² ? 4(-3)² = 12
This shows point (-4, -3) lies on hyperbola (1).
So, it?s normal to hyperbola.
Thus, it is evident that straight line (3) is normal to hyperbola (1); co-ordinate foot is (-4, -3).

We have, x = 2t³ ? 12t + 11  ???.(1)
Let v and f be velocity and acceleration respectively of particle at time t seconds.
Then, v = dx/dt = d(2t³ ? 12t + 11)/dt
= 6t² ? 12  ???.(2)
And f = dv/dt = d(6t² ? 12)/dt
= 12t   ???.(3)
Putting value of t = 2 in (3),
Therefore, displacement of particle at end of 2 seconds,
12t = 12(2)
= 24 cm/sec²

We have, x = a + bt + ct²  ???.(1)
Let, v and f be velocity and acceleration of a particle at time t seconds.
Then, v = dx/dt = d(a + bt + ct²)/dt = b + ct   ???.(2)
And f = dv/dt = d(b + ct)/dt = c  ???.(3)
Clearly, when c < 0,implies f < 0.
Hence in this case particle moves withan uniform retardation.

Let, particle moving with a uniform acceleration of f cm/sec².
By question initial velocity of particle is u = 10 cm/sec and its velocity after 8 seconds = v = 34 cm/sec.
Thus, using formula v = u + ft we get
34 = 10 + f*8
Or 8f = 24
Or f = 3
Therefore, required acceleration of particle is 3 cm/sec².
Thus, space described by particle in 8 seconds,
= [10*8 + 1/2(3)(8*8) [using formula s = ut +1/2(ft²)]
= 80 + 96
= 176 cm.