Ncert Exercises & Solutions

A particle executes the motion described by $x (t) = x_{0} (1 - e^{- γt}); t \geq 0 x_{0} > 0 .$ $x (t) = x_{0} (1 - e^{- γt}); t \geq 0 x_{0} > 0 .$ .

(a) Where does the particle start and with what velocity?

(b) Find maximum and minimum values of x(t), v{t), a(t). Show that x(t) and a(t) increase with time and v(t) decreases with time.

(a)

Hint: Particle starts its motion at t = 0.

Step 1: Put t = 0 in x(t) to calculate initial position of the particle.

Given,

$\begin{array}{l} x (t) = x_{0} (1 - e^{- γt}) \end{array}$

$When t = 0; x (t) = x_{0} (1 - e^{- 0}) = x_{0} (1 - 1) = 0 .$

Step 2: Calculate the velocity, v(t) = $\frac{dx}{dt}$

$v (t) = \frac{dx (t)}{dt} = x_{0} {γe}^{- γt} .$

Step 3: Calculate initial velocity by putting t = 0

V(0) = $x_{0} {γe}^{0} = x_{0} γ .$

(b)Hint: If f is increasing $\frac{df}{dt}$ > 0, if decreasing $\frac{df}{dt}$ < 0.

Step 1: Check weather f is increasing or decreasing

For x(t)

As here $\frac{dx}{dt}$ = $x_{0} {γe}^{- γt}$ > 0

i.e., x is increasing

For velocity, v(t)

$\frac{d v (t)}{d t} = - x_{0} γ^{2} e^{- γ t} < 0$

i.e., v(t) is decreasing

For acceleration, a(t)

$\frac{d a (t)}{d t} = x_{0} γ^{3} e^{- γ t} > 0$

i.e., a(t) is increasing

Step 2: Calculate maximum and minimum values

For increasing, the maximum value is at t = $\infty$ and minimum value is at t = 0,

For decreasing, the maximum value is at t = 0 and the minimum value is at t = $\infty$ .

For x(t)

As it is increasing

So,

$\begin{array}{ll} x (t) is maximum when t = \infty; x (t)]_{\max} = x_{0} \\ x (t) is minimum when t = 0; x (t)]_{\min} = 0 \\ v (t) i s d e c r e a \sin g \\ S o, \\ v (t) is maximum when t = 0; v (0) = x_{0} γ \\ v (t) is minimum when t = \infty; v (\infty) = 0 \\ a (t) i s i n c r e a \sin g \\ S o, \\ a (t) is maximum when t = \infty; a (\infty) = 0 \\ a (t) is minimum when t = 0; a (0) = - x_{0} γ^{2} \end{array}$

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