Ncert Exercises & Solutions

For the one-dimensional motion, described by

x = t - \sin t,

$x = t - \sin t,$ the following statements are given.

(a)	$x(t)>0$ for all $t>0$
(b)	$v(t)>0$ for all $t>0$
(c)	$a(t)>0$ for all $t>0$
(d)	$v(t)$ lies between $0$ and $2$

Choose the correct option:
1. (a), (c)
2. (b), (c)
3. (a), (d)
4. (b), (d)

(3) Hint: The first derivative of x gives velocity and the first derivative of velocity gives acceleration.

Step 1: Find the velocity and acceleration.

Given,

$\begin{array}{l} x = t - \sin t \\ velocity v = \frac{dx}{dt} = \frac{d}{dt} [t - \sin t] \\ = 1 - \cos t \\ Acceleration a = \frac{dv}{dt} = \frac{d}{dt} [1 - \cos t] = \sin t \end{array}$ $\begin{array}{l} x = t - \sin t \\ velocity v = \frac{dx}{dt} = \frac{d}{dt} [t - \sin t] \\ = 1 - \cos t \\ Acceleration a = \frac{dv}{dt} = \frac{d}{dt} [1 - \cos t] = \sin t \end{array}$

Step 2: Put the different values of t to find the correct answer.

As acceleration a > 0 for all t > 0
Hence, x(t) > 0 for all t > 0
Velocity v= 1 - cos t
When cos t = 1, velocity v = 0

$\begin{array}{l} V_{\max} = 1 - (\cos t)_{\min} = 1 - (- 1) = 2 \\ v_{\min} = 1 - (\cos t)_{\max} = 1 - 1 = 0 \end{array}$ $\begin{array}{l} V_{\max} = 1 - (\cos t)_{\min} = 1 - (- 1) = 2 \\ v_{\min} = 1 - (\cos t)_{\max} = 1 - 1 = 0 \end{array}$

Hence, v lies between 0 and 2.

$Acceleration a = \frac{dv}{dt} = - \sin t$ $Acceleration a = \frac{dv}{dt} = - \sin t$

$\begin{array}{l} When t = 0; x = 0, x = + 1, a = 0 \\ When t = \frac{π}{2}; x = 1, v = 0, a = - 1 \\ When t = π; x = 0, x = - 1, a = 1 \\ When t = 2 π; x = 0, x = 0, a = 0 \end{array}$ $\begin{array}{l} When t = 0; x = 0, x = + 1, a = 0 \\ When t = \frac{π}{2}; x = 1, v = 0, a = - 1 \\ When t = π; x = 0, x = - 1, a = 1 \\ When t = 2 π; x = 0, x = 0, a = 0 \end{array}$

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