MOTION IN A STRAIGHT LINE

VELOCITY

The rate of change of displacement (or position) w.r.t time is called velocity. Thus

$v=\frac{ds}{dt}$

ACCELERATION

The rate of change of velocity w.r.t time is called acceleration. Thus at time t

$a=\frac{dv}{dt}=\frac{d}{dt}\left(\frac{ds}{dt}\right)=\frac{d^{2}s}{d{t}^2}$

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.

Illustration 1 A particle moving in straight line covers s meter in t sec. If $s=3t+t^3$ then calculate the following :

i Acceleration after 2 sec.

ii Time when velocity is 51 m/s.

iii Distance travelled in third sec.

Solution $s=3t+t^3$

$velocity=\frac{ds}{dt}=3+3t^2$

$Acceleration=\frac{d^{2}s}{d{t}^2}=6t$

i Acceleration after 2 sec. = 6 x 2 = 12 m/s.

ii $velocity=\frac{ds}{dt}=51m/s$

$3+3t^2=51$$t^2=16$

$t=4$ sec.$\therefore$ After 4 sec the velocity is 51 m/s

iii $s=3t+t^3$

Distance after 3 sec = $3\times 3+3^3=36$ m

Distance after 2 sec = $3\times 2+2^3=14m$

$\therefore$ Distance travelled in third sec. = 36 – 14 = 22 m

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 2 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.

$\therefore$ distance covered in 8 sec. is 0.

$\therefore$ s = ut – 4.9 t² . . . (i)

or $0=u\times 8-4.9\times 8^2$

or $u=\frac{4.9\times 8\times 8}{8}=39.2m/s$

Differentiating,

$\frac{ds}{dt}=u-4.9\times 2t$

or $v=\frac{ds}{dt}=39.2-9.8t$

$\therefore$ velocity at t = 0 $=39.2-9.8\times 0=39.2$m/s

velocity at t = 2 $=39.2-9.8\times 2=19.6$ m/s.

$\frac{dy}{dx}$ AS A Rate MEASURER

Let y = f (x) be a single valued function of x. If increment in value of x is $\Delta$x and that of in y is $\Delta$y. Then the average rate of change of y w. r. t x in the interval (x, x + $\Delta$x) is the ratio $\frac{\Delta y}{\Delta x}$ (if it exists) is called the instantaneous rate of change of y w. r. t x at the point x.

$\underset{\Delta x\to 0}{\lim }\frac{\Delta y}{\Delta x}=\frac{dy}{dx}$

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 3: 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

$3x^2=12\frac{dy}{dx}$ $\Rightarrow$$\frac{dx}{dy}=\frac{12}{3x^2}$;;;;;In the interval at which the abscissa changes at a faster rate than the ;;;;;;;;;;ordinate we must have ;;;;;$\Rightarrow \left|\frac{dx}{dy}\right|>1or\left|\frac{12}{3x^2}\right|>1$ $\Rightarrow$$x\in (-2,2)$

Illustration 4 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 =\dfrac{dx}{dt}=2\m/s

since $\Delta$ABS and $\Delta$PQS are similar

$\therefore \frac{AB}{PQ}=\frac{BS}{QS}$ . . . (ii)

or $\frac{2}{3}=\frac{y}{y+x}$

or 3y = 2y + 2x

or y = 2x

and $\frac{dy}{dx}=2$ . . . (iii)

The rate at which the shadow is increasing $=\frac{dy}{dt}=\frac{dy}{dt}\cdot \frac{dx}{dt}$

substituting the values from (i) and (iii)

\dfrac{dy}{dt}=2\times 2=4\m/s

2 If SQ = l then we have to find $\frac{dl}{dt}$.

From (ii), $\frac{AB}{PQ}=\frac{BS}{QS}$

$\frac{2}{3}=\frac{\ell -X}{\ell }$

or $2\ell =3\ell -3x$

or $\ell$= 3x

or $\frac{d\ell }{dx}=3$

now $\frac{d\ell }{dt}=\frac{d\ell }{dx}\cdot \frac{dx}{dt}$

From (i) and (iv),

\dfrac{d\ell }{dt}=3\times 2=6\m/s