Motion In A Straight Line - Class 11 Medical Physics - Extra Questions

It is given that $$a=2t+4{{t}^{2}}$$

$$\because a=\dfrac{dv}{dt}$$

$$dv=2t+4{{t}^{2}}\,dt$$

$$\int{dv=\int{(2t+4{{t}^{2}}\,)dt}}$$

$$v={{t}^{2}}+\dfrac{4}{3}{{t}^{3}}$$

$$\because v=\dfrac{dx}{dt}$$

$$dx=({{t}^{2}}+\dfrac{4}{3}{{t}^{3}})dt$$

$$\int{dx=}\int\limits_{0}^{4}{\left( {{t}^{2}}+\dfrac{4}{3}{{t}^{3}} \right)}dt$$

$$x={{\left. \dfrac{{{t}^{3}}}{3}+\dfrac{{{t}^{4}}}{3} \right|}_{0}}^{4}$$

$$x=\dfrac{1}{3}\left[ {{4}^{3}}+{{4}^{4}} \right]$$

$$x=\dfrac{1}{3}(64+256)$$

$$x=106.66\,m$$

 

V=U+AT
Since v=0, we substitute values.
0=5-2t(because it is decelerating and hence it is negative).
T=2.5s
Displacement in nth second is given as
Sn=u+a(2n-1)/2

To find $$a$$ , we apply the relation $$a=\dfrac{dv}{dt}$$ , which is the same as $$a=\dfrac{\triangle v}{\triangle t}$$ over time intervals which $$v$$ varies linearly with $$t$$ . For the time interval between $$t_0=0$$ and $$t_1=2s$$ , the slope of the graph is positive and constant, equal to 
$${ a }_{ 1 }\dfrac { \triangle v }{ \triangle t } =\dfrac { \left( { v }_{ 1 }-{ v }_{ 0 } \right)  }{ \left( { t }_{ 1 }-{ t }_{ 0 } \right)  } =\dfrac { \left( { 2ms }^{ -1 }-0 \right)  }{ 2s-0 } =1{ ms }^{ -2 }$$
Between $$t_1=2s$$ and $$t_2=4s$$ , the slope is zero, so the acceleration is zero for this time interval. Between $$t_2=4s$$ and $$t_3=5s$$ , the slope is negative and constant, equal to
$${ a }_{ 2 }\dfrac { \triangle v }{ \triangle t } =\dfrac { \left( { v }_{ 3 }-{ v }_{ 2 } \right)  }{ \left( { t }_{ 3 }-{ t }_{ 2 } \right)  } =\dfrac { \left( { 0- }{ 2ms }^{ -1 } \right)  }{ 5s-4s } =-2{ ms }^{ -2 }$$
Figure 4.157 is graph of the instantaneous acceleration for the motion.

$$SI$$ units are understood in the $$y$$ and $$v$$ graphs shown. The acceleration graph is a horizontal line at $$–9.8\, m/s^{2}$$ .

We know that acceleration = rate of change of velocity/ time taken
Velocity in the first $$2$$ seconds is $$33.6\ km/h-30\ km/h=3.6\ km/h$$
$$\therefore$$ Acceleration $$=3.6/2=1.8\ km/h^{2}$$
Converting $$km/h^{2}$$ to $$m/s^{2}$$
$$1.8\times 1000/60\times 60=0.5\ m/s^{2}$$ .
Velocity after $$2$$ seconds is $$37.2\ km/h-33.6\ km/h=3.6\ km/h$$
$$\therefore$$ Acceleration $$=3.6/2=1.8\ km/h^{2}$$
Converting $$km/h^{2}$$ to $$m/s^{2}$$
$$1.8\times 1000/60\times 60=0.5\ m/s^{2}$$ .
Comparing acceleration from both instances it can be said that the acceleration is uniform.

A graph drawn by taking time along the x-axis and acceleration along the y-axis is called a velocity-time graph.

In the reference frame fixed to the train, the distance between the two events is obviously equal to $$l$$ . Suppose the train starts moving at time $$t=0$$ in the positive x direction and take the origin ( $$x=0$$ ) at the head-light of the train at $$t=0$$ . Then the coordinate of first event in the earth's frame is

$$\displaystyle x_1=\frac{1}{2}wt^2$$

and similarly the coordinate of the second event is

$$\displaystyle x_2=\frac{1}{2}w{(t+\tau)}^2-l$$

The distance between the two events is obviously,

$$\displaystyle x_1-x_2=l-w\tau\left(t+\frac{\tau}{2}\right)=0.242\:km$$

in the reference frame fixed on the earth.

For the two events to occur at the same point in the reference frame $$K$$ , moving with constant velocity $$V$$ relative to the earth, the distance travelled by the frame in the time interval $$T$$ must be equal to the above distance.

Thus $$\displaystyle V\tau=l-w\tau\left(t+\frac{\tau}{2}\right)$$

So, $$\displaystyle V=\frac{1}{\tau}-w\tau\left(t+\frac{\tau}{2}\right)=4.03\:m/s$$

The frame $$K$$ must clearly be moving in a direction opposite to the train so that if (for example) the origin of the frame coincides with the point $$x_1$$ on the earth at time $$t$$ , it coincides with the point $$x_2$$ at time $$t+\tau$$ .

$$time\quad to\quad reach\quad max\quad speed\quad from\quad 0=\quad \dfrac { 2.5 }{ 0.2 } =12.5s\\ displacement\quad while\quad achieving\quad max\quad speed\quad =\quad \dfrac { 0.2\times { 12.5 }^{ 2 } }{ 2 } =15.625m\\ So,\quad it\quad doesnt\quad reach\quad the\quad max\quad speed.\\ let\quad the\quad highest\quad speeed\quad achived\quad by\quad the\quad elevator\quad =\quad v\\ it\quad takes\quad time\quad t\quad to\quad reach\quad that\quad speed.\quad To\quad return\quad to\quad rest\quad it\quad must\quad \\ take\quad time\quad 2t.\quad total\quad displacement\quad during\quad this\quad time\quad =\quad \dfrac { 3 }{ 2 } vt\\ v=0.2t\\ and\quad 14\quad =\quad \dfrac { 3 }{ 2 } vt\Rightarrow 14=\dfrac { 3 }{ 2 } 0.2{ t }^{ 2 }\Rightarrow t=6.83\\ K=3t=20.5\Rightarrow 2K=41$$

$$\dfrac{x_0-2}{100}=\dfrac{4-x_0}{50} \rightarrow x_0=\dfrac{10}{3}m$$

Time interval	a-t graph	v-t graph	x-t graph
0<?<6s a, = 1	Straight line parallel with time axis and above time axis.	Straight line with (+) slope passing through origin	Parabola open up and passing through origin
6s<t< $$t_1$$	Straight line parallel with time axis and above time axis.	Straight line with (+) slope	Parabola open up
$$t_1 < t <t_2$$ Constant speed	Straight line parallel along time axis	Straight line parallel with time axis	Straight line with (+) slope
$$t_2 < t <40s$$ Retardation	Straight line parallel with time axis and below time axis.	Straight line with (-) slope	Parabola open down

The velocity of the particle at t = 4s can be given as
$$\vec{v}_4 = \vec{v}_0 + \Delta \vec{v}$$      ...(i)
where $$\Delta \vec{v}$$ = A

Acceleration-time graph is as follows:
We know that the slope of the velocity-time graph is equal to acceleration. The slope at A is maximum, decreases a, linearly, and becomes zero at C and then starts A increasing in the negative side and becomes maximum negative at E.
So at A, acceleration is maximum, at C zero, and at Is, acceleration is negative maximum.
Acceleration is positive and decreasing in magnitude. Also, the magnitude of velocity is decreasing, because velocity and acceleration are in opposite direction.

If car is starting from rest then its initial velocity is $$0$$

so $$u = 0$$

$$a = 2$$

$$t= 10$$

 

2) Using relation $$v= u+at$$

$$v= 0 + 2×10$$

$$v = 20 m/s$$

 

 

1) Using relation

 

$$v^2 - u^2 = 2as$$

$$(20)^2- 0 = 2 × 2 × s$$

$$s= 400 / 4$$

$$s = 100 m$$

 

 

 

(a) The $$x$$ and $$y$$ equations combine to give us the expression for $$\vec{r}$$ :
     $$\vec{r} = 18.0t\hat{i} + (4.00t − 4.90t^{2} )\hat{j}$$ , where $$\vec{r}$$ is in meters and $$t$$ is in seconds.
(b) We differentiate the expression for $$\vec{r}$$ with respect to time
     $$\vec{v}=\frac{d\vec{r}}{dt}=\frac{d}{dt}[18.0t\hat{i}+(4.00t-4.90t^{2})\hat{j}]$$
       $$=\frac{d}{dt}(18.0t)\hat{i}+\frac{d}{dt}(4.00t-4.90t^{2})\hat{j}$$
     $$\vec{v}=18.0\hat{i}+[4.00-(9.80)t]\hat{j}$$ , where $$\vec{v}$$ is in meters per second and $$t$$ is in seconds.
(c) We differentiate the expression for $$\vec{v}$$ with respect to time:
    $$\vec{a}=\frac{d\vec{v}}{dt}=\frac{d}{dt}[18.0\hat{i}+(4.00-(9.80)t)\hat{j}]$$
       $$\frac{d}{dt}(18.0)\hat{i}+\frac{d}{dt}[4.00-(9.80)t]\hat{j}$$
     $$\vec{a}=-9.80\hat{j}m/s^{2}$$
(d) By substitution,
      $$\vec{r}(3.00 s) = (54.0 m)\hat{i} − (32.l m)\hat{j}$$
      $$\vec{v}(3.00 s) = (18.0 m/s)\hat{i} − (25.4 m/s)\hat{j}$$
      $$\vec{a}(3.00s) =  (−9.80 m/s^{2})\hat{j}$$

(a) From $$x = −5.00\sin\omega t$$ , we determine the components of the velocity by taking the time derivatives of $$x$$ and $$y$$ :
   $$v_{x}=\frac{dx}{dt}=\left ( \frac{d}{dt} \right )(-5.00\sin\omega t)=-5.00\omega\cos t$$
and $$v_{y}=\frac{dy}{dt}=\left ( \frac{d}{dt} \right )(4.00-5.00\cos\omega t)=0+5.00\omega\sin t$$
At $$t=0$$
   $$\vec{v}=(-5.00\omega\cos0)\hat{i}+(5.00\omega\sin 0)\hat{j}=-5.00\omega\hat{i}m/s$$
(b) Acceleration is the time derivative of the velocity, so
   $$a_{x}=\frac{dv_{x}}{dt}=\frac{d}{dt}(-5.00\omega \cos\omega t)=+5.00\omega^{2}\sin\omega t$$
and $$a_{y}=\frac{dv_{y}}{dt}=\left ( \frac{d}{dt} \right )(5.00\omega\sin\omega t)=5.00\omega^{2}\cos\omega t$$
At $$t=0$$
    $$\vec{a}=(5.00\omega^{2}\sin o)\hat{i}+(5.00\omega^{2}\cos0)\hat{j}=5.00\omega^{2}\hat{j}m/s^{2}$$
(c) $$\vec{r}=x\hat{i}+y\hat{j}=(4.00m)\hat{j}+(5.00m)(-\sin\omega t\hat{i}-\cos\omega t\hat{j})$$
    $$\vec{v}=(5.00m)\omega[-\cos\omega t\hat{i}+\sin\omega t\hat{j}]$$
    $$\vec{a}=(5.00m)\omega^{2}[-\sin\omega t\hat{i}+\cos\omega t\hat{j}]$$
(d) The object moves in a circle of radius $$5.00 m$$ centered at $$(0, 4.00 m)$$