Kinematics Problems

1. The displacement of an object is given by the following equation.
$x=2t+t^{2}_{}-3t^{3}$

Calculate the velocity of the object when its acceleration is zero.

We have, $x=2t+t^{2}_{}-3t^{3}$

Or, velocity $v=\frac{dx}{dt}=\frac{d}{dt}(2t+t^{2}_{}-3t^{3})$

Or, $v=2+2t-9t^{2}$\ldots \ldots \ldots \ldots \ldots (1)

Acceleration, $a=$$\frac{dv}{dt}=\frac{d}{dt}(2+2t-9t^{2})$

Or, $a=2-18t$

Or, $2-18t=0$, as acceleration is zero

Or, $t=2/18=1/9$

Putting this value in equation (1), we get,

$v=2+2\times \frac{1}{9}-9(1/9)^{2}$

Or, $v=2+\frac{2}{9}-\frac{1}{9}$

Or, $v=19/9$.

2. An object is released from a certain height under the influence of gravity. State the nature of the velocity-displacement graph.

For an object undergoing free fall, the equation of motion is given by,

$x=ut+1/2 gt^{2}=1/2 gt^{2}$, initial velocity u = 0

velocity $v=\frac{dx}{dt}=\frac{d}{dt}(1/2 gt^{2})$

Or, $v=gt$

Or, $t=v/g$

Putting the value of t in the above equation, we get,

$x=1/2 g \frac{v^{2}}{g^{2}}$

Or, $v^{2}=2gx$

This is an equation of parabola. Therefore, the velocity-displacement graph gives a parabolic path.

3. An object is released from a height H. During the journey, it covers $H/2$distance in time $t_{1}$ and the remaining distance in time $t_{2}$. Prove that $t_{1}$$>$ $t_{2}$.

For the first half, we have the equations of motion,

$H/2=1/2 gt_{1}^{2}$ ……………. (1)

$v=gt_{1}^{}$ ……………… (2)

For the second half,

$H/2=vt_{2}+1/2 gt_{2}^{2}$

Or, $H/2=gt_{1}t_{2}+1/2 gt_{2}^{2}$ …………….. (3)

Comparing equations (1) and (3), we get,

$1/2 gt_{1}^{2}=gt_{1}t_{2}+1/2 gt_{2}^{2}$

Or, $t_{1}^{2}=2 t_{1}t_{2}+t_{2}^{2}$

Or, $2 t_{1}t_{2}+t_{2}^{2}-t_{1}^{2}=0$

Applying the formula of quadratic equation, we get,

<math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>t</mi><mn>2</mn></msub><mo>=</mo><mfrac><mrow><mo>-</mo><mn>2</mn><msub><mi>t</mi><mn>1</mn></msub><mo>+</mo><msqrt><mn>4</mn><msubsup><mi>t</mi><mn>1</mn><mn>2</mn></msubsup><mo>-</mo><mn>4</mn><mfenced><mrow><mo>-</mo><msubsup><mi>t</mi><mn>1</mn><mn>2</mn></msubsup></mrow></mfenced></msqrt></mrow><mn>2</mn></mfrac><mspace linebreak="newline"/><mo>&#xA0;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mo>=</mo><mfrac><mrow><mo>-</mo><mn>2</mn><msub><mi>t</mi><mn>1</mn></msub><mo>+</mo><mn>2</mn><msub><mi>t</mi><mn>1</mn></msub><msqrt><mn>2</mn></msqrt></mrow><mn>2</mn></mfrac><mspace linebreak="newline"/><mo>&#xA0;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mo>=</mo><mo>&#xA0;</mo><mo>-</mo><msub><mi>t</mi><mn>1</mn></msub><mo>+</mo><mn>1</mn><mo>.</mo><mn>44</mn><msub><mi>t</mi><mn>1</mn></msub><mspace linebreak="newline"/><mo>&#xA0;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mo>=</mo><mo>&#xA0;</mo><mn>0</mn><mo>.</mo><mn>44</mn><mo>&#xA0;</mo><msub><mi>t</mi><mn>1</mn></msub></math>

Therefore, $t_{1}>t_{2}$.

4. A stone of mass 0.01kg is released from the terrace of a building of height 70 m. Simultaneously another stone of mass 0.008kg is thrown up from the same place with a speed of 30 m/s. Calculate their relative speed.

For stone of mass 0.01kg, we apply kinematic equation,

$v_{1}=0-gt=-gt$

For a stone of mass 0.008kg, we apply the kinematic equation

$v_{2}=30-gt$

Relative velocity of first stone with respect to second stone,

$v_{12}=-gt-(30-gt)=-30$m/s

Similarly, $v_{21}=30-gt-(-gt)=30$m/s

So, the relative velocity is 30 m/s.

5. A racing car moves along a circular path of radius R at speed v so that it makes one revolution in every T seconds. If the car moves around a different circular path of radius 4R so that it takes one revolution in every T/2 seconds, then find the ratio of the magnitude of centripetal acceleration.

We have centripetal acceleration, $a=V^{2}/R$

Now, $V=2\pi R/T$

So, for the first track, centripetal acceleration $a_{1}^{}=\frac{4\pi ^{2}R^{2}}{RT^{2}}=\frac{4\pi ^{2}R^{}}{T^{2}}$

For the second track, $V=2\pi 4R/(T/2)=16\pi R/T$

For the second track, centripetal acceleration $a_{2}^{}=\frac{16\times
16\pi ^{2}R^{2}}{4RT^{2}}$

Or, $a_{2}^{}=\frac{64\pi ^{2}R^{}}{T^{2}}=16 a_{1}$

Or, $a_{1}$/$a_{2}^{}$=1/16

Or, $a_{1}:a_{2}=1:16$.

6. A bus starts from rest and accelerates uniformly over a time of 6 seconds for a distance of 110 m. Calculate the acceleration of the car.

We have, initial velocity, $u=0$, $t=6s$, $s=110m$, $a=?$

Using kinematic equation,

$s=ut+1/2 at^{2}$

Or, $110=0+1/2 (a\times 36)$

Or, $36a=220$

Or, $a=6.11 m/s^{2}$.

7. Four cyclists A, B, C and D are in motion. Their relative velocities are given below:

\vec{v}_{DC} = 20 m/s towards north

\vec{v}_{BC} = 20 m/s towards east

\vec{v}_{BA} = 20 m/s towards south

Calculate \vec{v}_{DA}.

We have,

<math xmlns="http://www.w3.org/1998/Math/MathML"><mover><msub><mi>v</mi><mrow><mi>D</mi><mi>C</mi></mrow></msub><mo>&#x2192;</mo></mover><mo>=</mo><mover><msub><mi>v</mi><mi>D</mi></msub><mo>&#x2192;</mo></mover><mo>-</mo><mover><msub><mi>v</mi><mi>C</mi></msub><mo>&#x2192;</mo></mover><mo>=</mo><mn>20</mn><mo>&#xA0;</mo><mover><mi>j</mi><mo>^</mo></mover></math>        ……….(1)

<math xmlns="http://www.w3.org/1998/Math/MathML"><mover><msub><mi>v</mi><mrow><mi>B</mi><mi>C</mi></mrow></msub><mo>&#x2192;</mo></mover><mo>=</mo><mover><msub><mi>v</mi><mi>B</mi></msub><mo>&#x2192;</mo></mover><mo>-</mo><mover><msub><mi>v</mi><mi>C</mi></msub><mo>&#x2192;</mo></mover><mo>=</mo><mn>20</mn><mo>&#xA0;</mo><mover><mi>i</mi><mo>^</mo></mover></math> …………..(2)

<math xmlns="http://www.w3.org/1998/Math/MathML"><mover><msub><mi>v</mi><mrow><mi>B</mi><mi>A</mi></mrow></msub><mo>&#x2192;</mo></mover><mo>=</mo><mover><msub><mi>v</mi><mi>B</mi></msub><mo>&#x2192;</mo></mover><mo>-</mo><mover><msub><mi>v</mi><mi>A</mi></msub><mo>&#x2192;</mo></mover><mo>=</mo><mo>-</mo><mn>20</mn><mo>&#xA0;</mo><mover><mi>j</mi><mo>^</mo></mover></math> ………….(3)

Subtracting equation (2) from equation (1), we get,

<math xmlns="http://www.w3.org/1998/Math/MathML"><mover><msub><mi>v</mi><mi>D</mi></msub><mo>&#x2192;</mo></mover><mo>-</mo><mover><msub><mi>v</mi><mi>B</mi></msub><mo>&#x2192;</mo></mover><mo>=</mo><mn>20</mn><mo>&#xA0;</mo><mover><mi>j</mi><mo>^</mo></mover><mo>-</mo><mn>20</mn><mo>&#xA0;</mo><mover><mi>i</mi><mo>^</mo></mover></math> ……………..(4)

Adding equation (3) and (4), we get

<math xmlns="http://www.w3.org/1998/Math/MathML"><mover><msub><mi>v</mi><mi>D</mi></msub><mo>&#x2192;</mo></mover><mo>-</mo><mover><msub><mi>v</mi><mi>A</mi></msub><mo>&#x2192;</mo></mover><mo>=</mo><mo>-</mo><mn>20</mn><mo>&#xA0;</mo><mover><mi>i</mi><mo>^</mo></mover><mspace linebreak="newline"/></math>

Or, \vec{v}_{DA} = $-20{i}$

So, \vec{v}_{DA} = 20 m/s towards west.

8. The displacement (in meter) of an object moving along x-axis is given by $x=9t+4 t^{2}$. Calculate Instantaneous velocity at $t=2s$.

$v=dx/dt=\frac{d}{dt}($$9t+4 t^{2}$)

Or, $v=9+8t$

Instantaneous velocity at $t=2s$is $v=9+8\times 2=25$m/s.

Please Share