Free Study Material

NCERT Solutions

NCERT Books

Revision Notes

Practice Problems

This is second of our many quizzes on Motion in one dimension. Click on this to access more such competitive quizzes on Mechanics.

Motion in one dimension Quiz 2

physics

 The velocity in the $x-t$ graph during $C-D$ is: $ \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/1a365eb3-1764-4d30-9c59-435d0a3220335367066987295560373.png"/> 

 Calculate slope at various instances to predict the nature of velocity in $x-t$ graph. Value of slope in the $x-t$ curve gives an idea of the nature of velocity at various time instances. 

  The slope of the position time graph is $\dfrac{\text{dx}}{dt} $ at any given instant $t$. This is Velocity of the object, as $ \\ V=\dfrac{\text{dx}}{dt}. \\ $ We can see from the graph that the slope is increasing between points C and D. Thus, the velocity is increasing between C and D.

 In an accelerating motion, the distance travelled by a particle in the last second ($n$) equal  to the distance travelled in the first two seconds. If the particle's initial velocity is $10m/s$ and $a=2m/s^2$, then the total journey time is:  

 Use second equation of motion to get displacement of the first two seconds, $s=ut+\dfrac{1}{2}a{{t}^{2}}$ and displacement only in the last second n is ${{s}_{n}}=u+\dfrac{a}{2}\left( 2n-1 \right)$. 

  Formula for Distance travelled in ${{\text{n}}^{\text{th}}}$ second $ \\{{S}_{n}}=u+\dfrac{a}{2}(2n-1), \\ $ $‘u’$ is initial velocity and a is acceleration of the body $ \\ $ Displacement travelled in the first two second, $ \\s=ut+\dfrac{1}{2}a{{t}^{2}}=10\times 2+\dfrac{1}{2}\times 2\times {{2}^{2}}=24 \\ $  This is equal to the distance travelled in the last second $n$ $\\ {{s}_{n}}=10+\dfrac{2}{2}(2n-1)=9+2n \\ $ As $s={{s}_{n}} $ $ \\24=9+2n \\ \therefore n=7.5, \\ $ So the total time of the journey is $7.5 sec$.

 A man is travelling with a velocity of $10\,m/s$ in the north direction, and his velocity increases by $2$ second. What is the total distance travelled in these two seconds?

Use the second equation of motion to get displacement as $S=ut+\dfrac{1}{2}a{{t}^{2}}$. 

 As velocity increases by $2 m/s$ in $2s$, the acceleration towards north is $ \\ \text{a=}\dfrac{\Delta \text{v}}{t}=\dfrac{2}{2}=1 \\ $ Second equation of motion is, $ \\s=ut+\dfrac{1}{2}a{{t}^{2}} \\ \therefore \,\,s=10\times 2+\dfrac{1}{2}\times 1\times {{2}^{2}}=22\,m$

If $x\,=\,3+{{t}^{2}}$ m, and $y\,=\,2t$ m,  then the speed at $3$ sec would be 

 Calculate ${{V}_{x}}$ and ${{V}_{y}}$ as $V=\dfrac{ds}{dt}$,  where $s$ is the position which can be along either $x$ or $y$ axis and then use $V=\sqrt{{{V}_{x}}^{2}+{{V}_{y}}^{2}}$ to get magnitude of velocity. 

 $x=3+{{t}^{2}} \\ \Rightarrow {{V}_{x}}=\dfrac{dx}{dt}=2t \\ \Rightarrow {{\left( {{V}_{x}} \right)}_{t=3}}=6\text{m/s} \\ \Rightarrow y=2t \\ \Rightarrow {{V}_{y}}=\dfrac{dy}{dt} \\ \Rightarrow {{V}_{y}}=2\text{m/s} \\ $ These components of velocity are mutually perpendicular. Then their resultant would be $ \\ \Rightarrow V=\sqrt{{{V}_{x}}^{2}+{{V}_{y}}^{2}}\\ \Rightarrow V =\sqrt{{{2}^{2}}+{{6}^{2}}}\\ \therefore V =\sqrt{40}\,\text{m/s}$

 What can we say about this curve? $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/263e9579-6da1-46fe-9975-3043970ba6681226484001572958703.png"/> 

 The acceleration is indicated by the slope of the velocity-time graph. Retardation occurs when acceleration is applied against the initial velocity.

 From the velocity time graph, we can see the slope  $\dfrac{\text{dV}}{dt}$ is constant and positive. As, acceleration $a=\dfrac{\text{dV}}{dt}$ so this is a graph where the acceleration is constant and positive. $ \\ $ Here, the initial velocity is negative at $t=0$ secs, which becomes zero and then positive with time. Thus, the acceleration acts against the initial velocity. This is a graph of retarded motion with a positive slope