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Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of 15 m s-1and 30 m s-1. Verify that the graph shown in Fig. 3.27 correctly represents the time variation of the relative position of the second stone with respect to the first. Neglect air resistance and assume that the stones do not rebound after hitting the ground. Take g = 10 m s-2. Give the equations for the linear and curved parts of the plot.


bold For bold space bold first bold space bold stone comma space

Initial space velocity comma space straight u subscript 1 equals space 15 space straight m divided by straight s
Acceleration comma space straight a space equals space – straight g space equals space – space 10 space straight m divided by straight s squared

Now comma space using space the space second space equation space of space motion comma space

straight x subscript 1 space equals space straight x subscript 0 space plus space straight u subscript 1 straight t space plus space left parenthesis 1 divided by 2 right parenthesis at squared

space straight x subscript 0 space equals space 200 space straight m space is space the space height space of space the space cliff. space

Therefore comma space

straight x subscript 1 space equals space 200 space plus space 15 straight t space minus space 5 straight t to the power of 2 space end exponent space space space space space space space... left parenthesis straight i right parenthesis space

When space stone space hits space the space ground comma space straight x subscript 1 space equals space 0

therefore space – space 5 straight t squared space plus space 15 straight t space plus space 200 space equals space 0
rightwards double arrow space space space space space space space straight t squared space – space 3 straight t space – space 40 space equals space 0
rightwards double arrow space space space straight t to the power of 2 space end exponent – space 8 straight t space plus space 5 straight t space – space 40 space equals space 0 space

rightwards double arrow space straight t space left parenthesis straight t space – space 8 right parenthesis space plus space 5 space left parenthesis straight t space – space 8 right parenthesis space equals space 0 space

straight i. straight e. comma space space space space space straight t space equals space 8 space straight s space or space straight t space equals space – space 5 space straight s space

straight t space cannot space be space negative space as space the space stone space is space
projected space at space straight a space certain space time. space

Therefore comma space straight t space equals space 8 space sec

bold italic F bold italic o bold italic r bold space bold italic s bold italic e bold italic c bold italic o bold italic n bold italic d bold space bold italic s bold italic t bold italic o bold italic n bold italic e comma space

Initial space velocity comma space straight u subscript 2 space equals space 30 space straight m divided by straight s space
Acceleration comma space straight a space equals space – space straight g space equals space – space 10 space straight m divided by straight s squared

Now comma space using space the space second space equation space of space motion comma space

straight x subscript 2 space end subscript equals space straight x subscript 0 space end subscript plus space straight u subscript 2 straight t space plus space left parenthesis 1 divided by 2 right parenthesis at squared space

space space space space equals space 200 space plus space 30 straight t space minus space 5 straight t squared space space space space space space space space space space space space space space space space space space space space... left parenthesis ii right parenthesis

When space stone space hits space the space ground comma space straight x subscript 2 space equals space 0

So comma space
space space space space space – space 5 straight t squared space plus space 30 space straight t space plus space 200 space equals space 0 space
rightwards double arrow space space straight t to the power of 2 space end exponent – space 6 straight t space – space 40 space equals space 0 space
rightwards double arrow space space straight t squared space – space 10 straight t space plus space 4 straight t space plus space 40 space equals space 0 space

rightwards double arrow space space space space straight t space left parenthesis straight t space – space 10 right parenthesis space plus space 4 space left parenthesis straight t space – space 10 right parenthesis space equals space 0 space

rightwards double arrow space space space space space space space straight t space left parenthesis straight t space – space 10 right parenthesis space left parenthesis straight t space plus space 4 right parenthesis space equals space 0

rightwards double arrow space space space space space straight t space equals space 10 space straight s space or space straight t space equals space – space 4 space straight s

straight t space cannot space be space negative space again. space

therefore space straight t space equals space 10 space sec space

On space subtracting space equations space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis comma space

straight x squared space minus space straight x to the power of 1 space equals space left parenthesis 200 space plus space 30 straight t space minus 5 straight t squared right parenthesis space minus space left parenthesis 200 space plus space 15 straight t space minus 5 straight t squared right parenthesis

straight x subscript 2 space minus space straight x subscript 1 space equals 15 straight t space space space space space space space space space space space space space space... space left parenthesis iii right parenthesis

Equation space left parenthesis iii right parenthesis space represents space the space linear space path space of space both space stones. space

Due space to space this space linear space relation space between space left parenthesis straight x subscript blank to the power of 2 space end exponent end subscript – space straight x subscript 1 right parenthesis space and space straight t comma
the space path space remains space straight a space straight space line space till space 8 space straight s. space

At space straight t equals space 8 space sec comma space stones space are space seperated space
by space straight a space maximum space distance.

left parenthesis straight x subscript 2 space end subscript – space straight x subscript 1 right parenthesis subscript max space end subscript equals space 15 cross times space 8 space equals space 120 space straight m

After space 8 space straight s comma space only space second space stone space is space in space motion space
whose space variation space with space time space is space given space by space the space
quadratic space equation colon

straight x to the power of 2 space end exponent – space straight x to the power of 1 space equals space 200 space plus space 30 straight t space – space 5 straight t squared

Therefore comma space the space equation space of space linear space and space
curved space path space is space given space by comma

straight x subscript 2 space end subscript – space straight x subscript 1 space end subscript equals space 15 straight t comma space represents space linear space path.

straight x subscript 2 space end subscript ­ – space straight x subscript 1 space end subscript equals space 200 space plus space 30 straight t space – space 5 straight t squared comma space represents space the space curved space path. 180 Views

The velocity-time graph of a particle in one-dimensional motion is shown in Fig. 3.29:

Which of the following formulae are correct for describing the motion of the particle over the time-interval t1 to t2 :

(a) x(t2 ) = x(t1) + v (t1) (t2 – t1) +(½) a (t2 – t1)

(b) v(t2 ) = v(t1) + a (t2 – t1

(c) vaverage = (x(t2) – x(t1))/(t2 – t1

(d) aaverage = (v(t2) – v(t1))/(t2 – t1

(e) x(t2 ) = x(t1) + vaverage (t2 – t1) + (½) aaverage (t2 –t1)

(f) x(t2 ) – x(t1) = area under the v-t curve bounded by the t-axis and the dotted line shown.


The correct formulae describing the motion of the particle are (c), (d) and, (f).

The given graph has a non-uniform slope.

Hence, the formulae given in (a), (b), and (e) cannot describe the motion of the particle.

Only relations given in (c), (d), and (f) are correct equations of motion.
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Figure 3.25 gives a speed-time graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude ? In which interval is the average speed greatest ? Choosing the positive direction as the constant direction of motion, give the signs of v and a in the three intervals. What are the accelerations at the points A, B, C and D?


Acceleration is given by the slope of the speed-time graph.

Since the slope of the given speed-time graph is maximum in interval 2, the average acceleration will be the greatest in this interval.

The height of the curve from the time-axis gives the average speed of the particle.
In the given graph, it is clear that the height is the greatest in interval 3.
Hence, the average speed of the particle is the greatest in interval 3.

In interval 1:

The slope of the speed-time graph is positive. Hence, acceleration is positive. Similarly, the speed of the particle is positive in this interval.

In interval 2:

The slope of the speed-time graph is negative. Hence, acceleration is negative in this interval. However, speed is positive because it is a scalar quantity.

In interval 3:

The slope of the speed-time graph is zero. Hence, acceleration is zero in this interval. However, here the particle acquires some uniform speed. It is positive in this interval.

Points A, B, C, and D are all parallel to the time-axis. Hence, the slope is zero at these points. Therefore, at points A, B, C, and D, acceleration of the particle is zero. 
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On a long horizontally moving belt (Fig. 3.26), a child runs to and fro with a speed 9 km h-1 (with respect to the belt) between his father and mother located 50 m apart on the moving belt. The belt moves with a speed of 4 km h-1. For an observer on a stationary platform outside, what is the

(a) speed of the child running in the direction of motion of the belt ?.

(b) speed of the child running opposite to the direction of motion of the belt ?

(c) time taken by the child in (a) and (b) ?

Which of the answers alter if motion is viewed by one of the parents ?


a) Speed of the belt, vB4 km/h

Speed of the boy, vb = 9 km/h

The boy is running in the same direction of the motion of the belt. Therefore, speed of the child is given by, 

vbB = vb + vB = 9 + 4 = 13 km/h

b) The boy is running in a direction opposite to the direction of the motion of the belt. 

Therefore, his speed as observed by the stationary observer is given by, 

vbB = vb + (– vB) = 9 – 4 = 5 km/h

c) Distance between child and his parents = 50 m 
Given that, both parents are standing on the moving belt. Therefore, the speed of the child in either direction as observed by the parents will remain the same i.e., 9 km/h = 2.5 m/s.

Hence, time taken by the child to move towards one of his parents = fraction numerator 50 over denominator 2.5 end fraction space equals space 20 space s e c

If the motion is viewed by any one of the parents, answers obtained in (a) and (b) get altered. This is because the child and his parents are standing on the same belt and hence, are equally affected by the motion of the belt. Therefore, for both parents (irrespective of the direction of motion) the speed of the child remains the same i.e., 9 km/h.

For this reason, it can be concluded that the time taken by the child to reach any one of his parents remains unaltered.
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The speed-time graph of a particle moving along a fixed direction is shown in Fig. 3.28. Obtain the distance traversed by the particle between (a) t = 0 s to 10 s, (b) t = 2 s to 6 s.
What is the average speed of the particle over the intervals in (a) and (b)?



a)

Distance travelled by the particle = Area under the given graph.

                                           = (1/2) × (10 - 0) × (12 - 0)

                                           = 60 m


Average space Speed space equals space Distance over Time equals 60 over 10 space equals space 6 space straight m divided by straight s
b) 

Let s1 and s2 be the distances covered by the particle between time = 2 s to 5 s and t = 5s to 6 s respectively.

Total distance (s) covered by the particle in time t = 2 s to 6 s is, 

                           s = s1 + s2                     … (i) 
For distance s1

Let u1 be the velocity of the particle after 2 s, and 
a1 be the acceleration of the particle in t = 0 to t = 5s. 

The particle undergoes uniform acceleration in the first 5 seconds. 

Therefore, using first equation of motion v = u+at, we have

    12 = 0 + a1 × 5

i.e.,  a1 = 12/5  = 2.4 m/s2

Again, using the first equation of motion, we have 

    v at 

      = 0 + 2.4 × 2

      = 4.8 m/s

Distance travelled by the particle between time 2 s and 5 s i.e., in 3 s

s1 = u1 t + (1/2) a1 t

   = 4.8 × 3 + (1/2) × 2.4 × (3)
 
   = 25.2 m                                   ...(ii) 


For distance s2

Let a2 be the acceleration of the particle between time t = 5 s and t = 10 s.

Using first equation of motion, 

v = u + at

0 = 12 + a2 × 5

a2 = -12 / 5 = - 2.4 ms-2

Therefore, 

Distance travelled by the particle in 1s (i.e., between = 5 s and t = 6 s),

s2u2 t + (1/2)a2 t2

   = 12 × 1 + (1/2) (-2.4) × (1)

   = 12 - 1.2 = 10.8 m                 ...(iii) 

Now, from equations (i), (ii) and (iii), we have

s = 25.2 + 10.8 = 36 m

∴ Average speed = 36 / 4 = 9 m/s 




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