Motion in a Straight Line

To find the average velocity given the motion of a person traveling along a straight line with two different velocities, we start by calculating the average velocity ($ v_{\text{avg}} $) using the total distance traveled divided by the total time taken.

The distances and velocities are given as follows:

Distance $ x $ at velocity $ v_1 = 5 \, \text{m/s} $

Distance $ \frac{3}{2}x $ at velocity $ v_2 $

The formula for average velocity is:

$ v_{\text{avg}} = \frac{\text{Total Distance}}{\text{Total Time}} $

Given:

$ v_{\text{avg}} = \frac{50}{7} \, \text{m/s} $

The total distance traveled is:

$ x + \frac{3x}{2} = \frac{5x}{2} $

The time taken to cover each segment is given by:

Time for $ x $: $ t_1 = \frac{x}{v_1} = \frac{x}{5} $

Time for $ \frac{3x}{2} $: $ t_2 = \frac{\frac{3x}{2}}{v_2} = \frac{3x}{2v_2} $

Now, using the formula for average velocity:

$ \frac{50}{7} = \frac{\frac{5x}{2}}{\frac{x}{5} + \frac{3x}{2v_2}} $

Simplifying the equation:

$ \frac{50}{7} = \frac{5/2}{\frac{1}{5} + \frac{3}{2v_2}} $

Cross-multiplying gives:

$ \frac{1}{5} + \frac{3}{2v_2} = \frac{7}{20} $

Solving for $ \frac{3}{2v_2} $:

$ \frac{3}{2v_2} = \frac{7}{20} - \frac{1}{5} = \frac{7-4}{20} = \frac{3}{20} $

Finally, solving for $ v_2 $:

$ \frac{3}{2v_2} = \frac{3}{20} $

$ v_2 = 10 \, \text{m/s} $

Thus, the value of $ v_2 $ is $ 10 \, \text{m/s} $.

Here, we will use the concept of relative motion.

Let initial (at $t = 0$) position of both cars is x = 0 (As they cross each other at t = 0 for Ist time)

given : For P

${a_P} \propto t$

For Q,

${a_Q} = a'$ (Let) (constant)

$ \Rightarrow {a_{{p^{(t)}}}} = kt$ where, k = constant

$ \Rightarrow {{d{v_p}(t)} \over {dt}} = kt$ (as $a = {{dv} \over {dt}}$)

$ \Rightarrow \int_0^{{v_p}(t)} {d{v_p}(t) = \int_0^t {ktdt} } $ [let p and Q both starts from rest]

$ \Rightarrow {v_p}(t) = {{k{t^2}} \over 2}$

$ \Rightarrow {{d{x_p}(t)} \over {dt}} = {{k{t^2}} \over 2}$ [As $v = {{dx} \over {dt}}$]

$ \Rightarrow \int_0^{{x_p}(t)} {d{x_{p(t)}} = \int_0^t {{{k{t^2}} \over 2}dt} } $

$ \Rightarrow {x_p}(t) = {{k{t^3}} \over 6}$ .... (1)

Now, for Q,

${x_Q}(t) = {1 \over 2}{a^1}{t^2}$ .... (2) [using Newton's 2nd equation of motion]

So, relative position of P w.r.t. Q,

${x_{P{Q^{(t)}}}} = {x_p}(t) - {x_Q}(t)$

$ \Rightarrow {x_{P{Q^{(t)}}}} = {{k{t^3}} \over 6} - {1 \over 2}{a^1}{t^2}$ (From (1) and (2))

When both cars cross each other, $X_{PQ}=O$

As $X_{PQ}(t)$ is a cubic polynomial, so it has maximum 3 roots.

Hence, the maximum number of crossing = 3

Given the displacement of the particle $x^2 = 1 + t^2$, we want to find the acceleration, which is the second derivative of displacement with respect to time, $a = \frac{d^2x}{dt^2}$, and we are given that the acceleration at any time $t$ is $x^{-n}$, for us to find the value of $n$.

First, let's find the first derivative of displacement with respect to time, which gives us the velocity. Differentiating $x^2 = 1 + t^2$ with respect to t, we get:

$2x\frac{dx}{dt} = 2t$

This simplifies to:

$\frac{dx}{dt} = \frac{t}{x}$

Now, let's differentiate this velocity to find the acceleration:

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

To differentiate $\frac{t}{x}$ with respect to $t$, we'll use the quotient rule:

$\frac{d}{dt}\left(\frac{t}{x}\right) = \frac{x\cdot 1 - t\cdot \frac{dx}{dt}}{x^2}$

Substitute $\frac{dx}{dt} = \frac{t}{x}$ into the equation:

$a = \frac{x(1) - t(\frac{t}{x})}{x^2} = \frac{x - \frac{t^2}{x}}{x^2} = \frac{x^2 - t^2}{x^3}$

Recalling that the displacement equation given was $x^2 = 1 + t^2$, substitute this into our expression for acceleration:

$a = \frac{1 + t^2 - t^2}{x^3} = \frac{1}{x^3}$

Thus, the acceleration of the particle at any time $t$ is $x^{-3}$, which means our value for $n$ is $3$.

Given that a body is moving on a frictionless plane and starts from rest, the motion can be assumed to be uniformly accelerated motion. The formula for the distance covered in uniformly accelerated motion from rest is given by $s = ut + \frac{1}{2}at^2$, where:

• $s$ is the distance covered,


• $u$ is the initial velocity (which is 0 since the body starts from rest),


• $a$ is the acceleration, and


• $t$ is the time.

Since the body starts from rest ($u=0$), the formula simplifies to $s = \frac{1}{2}at^2$.

The distance moved between $t = n - 1$ and $t = n$, denoted as $S_n$, can be found by calculating the distance covered by the end of time $n$ and subtracting the distance covered by the end of time $n-1$. Let's denote the total distance covered by time $n$ as $S(n)$, which according to the formula is $S(n) = \frac{1}{2}a n^2$. Thus, $S_n = S(n) - S(n-1)$.

Therefore, we have:

$S_n = \frac{1}{2}an^2 - \frac{1}{2}a(n-1)^2$

Simplifying this, we get:

$S_n = \frac{1}{2}a(n^2 - (n^2 - 2n + 1))$

This simplifies to:

$S_n = \frac{1}{2}a(2n - 1)$

Similarly,

$S_{n-1} = \frac{1}{2}a((n-1)^2 - (n-2)^2)$

Simplifying:

$S_{n-1} = \frac{1}{2}a((n-1)^2 - (n^2 - 4n + 4))$

Which further simplifies to:

$S_{n-1} = \frac{1}{2}a(2n - 3)$

Hence, the ratio $\frac{S_{n-1}}{S_n}$ can be calculated:

$\frac{S_{n-1}}{S_n} = \frac{\frac{1}{2}a(2n - 3)}{\frac{1}{2}a(2n - 1)} = \frac{2n - 3}{2n - 1}$

For $n = 10$,

$\frac{S_{n-1}}{S_n} = \frac{2(10) - 3}{2(10) - 1} = \frac{20 - 3}{20 - 1} = \frac{17}{19}$

Given that the ratio is represented as $\left(1 - \frac{2}{x}\right)$, we have:

$\frac{17}{19} = 1 - \frac{2}{x}$

Solving this equation for $x$ gives:

$1 - \frac{17}{19} = \frac{2}{x}$

$\frac{2}{19} = \frac{2}{x}$

Thus:

$x = 19$

A bus traveling along a straight highway at a speed of 72 km/h comes to a stop within 4 seconds after the brakes are applied. To find the distance the bus travels during this time (assuming uniform deceleration), follow these steps:

First, convert the initial speed from km/h to m/s:

$\begin{aligned} u &= 72 \times \frac{5}{18} = 20 \, \text{m/s} \end{aligned}$

Given:

Initial speed ($ u $) = 20 m/s

Final speed ($ v $) = 0 m/s

Time ($ t $) = 4 s

Using the first equation of motion:

$\begin{aligned} v &= u + at \\ 0 &= 20 + 4a \\ a &= -5 \, \text{m/s}^2 \end{aligned}$

Now, use the second equation of motion to find the distance traveled ($ S $):

$\begin{aligned} S &= ut + \frac{1}{2}at^2 \\ S &= 20 \times 4 + \frac{1}{2} \times (-5) \times 4^2 \\ S &= 80 - 40 \\ S &= 40 \, \text{m} \end{aligned}$

Therefore, the distance traveled by the bus during the time the brakes are applied is 40 meters.