Calculations
ref:
http://www.learningprocessing.com/examples/chapter-16/example-16-14/)
http://www.askiitians.com/iit_jee-Applications_of_Derivatives/Motion-in-a-Straight-Line
Motion in a straight line
Velocity
The rate of change of displacement (or position) w.r.t time is called velocity. Thus
v = ds/dt
Acceleration
The rate of change of velocity w.r.t time is called acceleration. Thus at time t
a = dv/dt = d/dt (ds/dt) = d2s/dt2
NOTE : The direction of acceleration is in the direction of velocity or opposite to it. When the direction of acceleration is opposite to the direction of velocity then it is called retardation. retardation means negative acceleration.
Example 1: A particle moving in straight line covers s meter in t sec. If then calculate the following :
¼i½ Acceleration after 2 sec.
¼ii½ Time when velocity is 51 m/s.
¼iii½ Distance travelled in third sec.
Solution: s = 3t + t3
velocity = ds/dt = 3 + 3t2
Acceleration = d2s/dt2 = 6t
¼i½ Acceleration after 2 sec. = 6 x 2 = 12 m/s.
¼ii½ velocity = ds/dt = 51 m/s
3 + 3t2 = 51
t2 = 16
t = 4 sec.
After 4 sec the velocity is 51 m/s
¼iii½
Distance after 3 sec = 3 × 3 + 33 = 36 m
Distance after 2 sec = 2 × 2 + 23 = 14 m
Distance travelled in third sec. = 36 – 14 = 22 m
Example 2: A particle moving in straight line covers s meter in t sec. If 1/2 sec.
Solution: s = 4t2 + t
v = ds/dt = 8t + 2
a = d2s/dt2 = 8 = constant
t = velocity in 1/2 = 8 × 1/2 + 2 = 6 sec
acceleration at t = 1/2 = 8 m/s
MOTION UNDER GRAVITY
When a particle is thrown vertically upwards then it comes back due to the gravity. This motion of a particle is due to the acceleration g. When the particle is going upward, the value of g is negative and when it is coming back, the value of g is positive. At maximum height the velocity of a particle is zero. The value of g is 9.8 m/s2 or 980 cm/s2.
Illustration 3: A ball is thrown upwards which comes back after 8 sec. on Earth. If the equation of motion is s = ut – 4.9 t2, where s is in meter and t in sec. then find the velocity at t = 0 and t = 2.
Solution : Since it comes back after 8 sec.
∴ distance covered in 8 sec. is 0.
∴ s = ut – 4.9 t2 . . . (i)
or 0 = u × 8 – 4.9 × 82
or u = 4.9 × 8 × 8 / 8 = 39.2 m/s
Differentiating,
or ds/dt = u – 4.9 × 2t
∴ v = ds/dt = 39.2 – 9.8t
velocity at t = 2 = 39.2 – 9.8 × 2 = 19.6 m/s.
Illustration 4: The equation of motion of a particle moving vertically is s = at2 + bt, where s is in meter and t in sec. If acceleration is –9.8 m@s2 and maximum height is 44 m then find the velocity after 1 sec.
Solution : Given s = at2 + bt . . . (i)
differentiating w.r.t. t
v = ds/dt = 2at + b
acceleration = dv/dt = t2s / dt = d2s / dt2 = 2a
∴ 2a = –9.8 => a = –4.9 m/s
at maximum height v = 0
∴ b – 9.8 t = 0
or t = b/9.8
∴ for maximum height
s = at2 + bt
= –4.9 (b/9.8)2 + b × b / 9.8
= b2/9.8 – b2/19.6 = b2/19.6
∴ 44 = b2 / 19.6
or b2 = 44 × 19.6 or b = 29.4
velocity after 1 sec.
v = 2a × 1 + b
= 2 × (–4.9) + 29.4 = 19.6 m/s
Illustration 5: A man standing on 12 m high pillar, throw a ball upwards. The equation of motion is s = 16.9 t – 4.9 t2, where s is in meter and t in sec. Find the time taken by the ball in upward motion. Also find the time taken by the ball when it comes back to earth.
Solution : s = 16.9 t – 4.9 t2
ds/dt = 16.9 – 9.8 t
=> v = 16.9 – 9.8t
Let at time t in upward motion, the velocity of the ball is zero.So
0 = 16.9 – 9.8 t
=> t = 16.9/9.8 = 169/98 sec
So the ball will take 169 / 98 sec in upward motion.
dy/dx AS A RATE MEASURER
Let y = f (x) be a single valued function of x. If increment in value of x is Δ x and that of in y is Δ y. Then the average rate of change of y w. r. t x in the interval (x, x + Δx) is the ratio (if it exists) is called the instantaneous rate of change of y w. r. t x at the point x.
Therefore the derivative of a function at a point x represents the rate measurer of y w. r. t x at the point x.
Illustration 6: On the curve x3=12y, find the interval at which the abscissa changes at a faster rate than the ordinate.
Solution: Given x3=12y, on differentiating w. r. t x
3x2 = 12 dy/dx => dx/dy = 12/3x2
In the interval at which the abscissa changes at a faster rate than the ordinate we must have
Illustration 7: A man of 2 m height moves away from a 3m high lamp post at the speed 2m/s. Find, at what rate (1) the length of his shadow increases. (2) the shadow is moving.
Solution : (1) Let PQ is the lamp, AB is the man, shadow BS = y m and the distance of the man from the foot of the lamp post BQ = x m.
speed (velocity) of the man = dx/dt = 2 . . . (i)
since ΔABS and ΔPQS are similar
AB / PQ = BS / QS . . . (ii)
or 2/3 = y/y+x
or 3y = 2y + 2x
or y = 2x
and dy/dx = 2 . . . (iii)
The rate at which the shadow is increasing dy/dt = dy/dt /dt
substituting the values from (i) and (iii)
dy/dt = 2 × 2 = 4 m/s
¼2½ If SQ = l then we have to find dl/dt.
From (ii), AB/PQ = BS/QS
2/3 = l – X / l
or 2l = 3l – 3x
or l = 3x
or dl/dx = 3
now dl / dt = dl / dx × dx/dt
From (i) and (iv),
dl / dt = 3 × 2 = 6 m/s
Illustration 8: The volume of a cube is increasing at a constant rate. Prove that the rate of change of surface is inversely proportional to the length of edge.
Solution : Let the edge of the cube is x, volume V and surface is S.
Now V = x3 and S = 6x2
dV/dx = 3x2 and dS / dx = 12x
Given dV/dt = constant = k
we know that dV/dt = dV/dt × dx/dt = k
3x2 × dx/dt = k or dx/dt = k 3x2
now, dS/dt = dS/dt × dx/dt
= 12 × k / 3x2 = 4k / x
dS / dt = 4k / x or dS/dt μ 1/x