The following questions numbered 1, 2 and 3 are based on the velocity-time graph of a particle in one dimensional motion shown in the adjoining figure.

**(1)**The displacement of the particle during the first second of its motion is nearly

(a) 0.25 m

(b) 0.5 m

(c) 1 m

(d) 1.5 m

(e) 2 m

The area under the velocity-time curve gives the displacement. The portion of the graph for the interval from zero to 1 second is straight and the area under the curve is triangular and is equal to (½)×1×4 = 2 m.

**(2)**The average accelerations during the 3

^{rd}second and 7

^{th}second respectively are nearly

(a) 0.5 ms

^{–2}**and 2 ms**^{–2}
(b) 1 ms

^{–2}**and –2 ms**^{–2}
(c) 1 ms

^{–2}**and 2 ms**^{–2}
(d) 1 ms

^{–2}**and –2 ms**^{–2}
(e)

**zero and –2 ms**^{–2}
The 3

^{rd}second is the interval from 2 seconds to 3 seconds and during this time the velocity increases from 7 ms^{–1}to 8 ms^{–1}. The acceleration is therefore (8 – 7)/1 = 1 ms^{–2}.
The 7

^{th}second is the interval from 6 seconds to 7 seconds and during this time the velocity*decreases*from 8 ms^{–1}to 6 ms^{–1}.
The acceleration is therefore (6 – 8)/1 = – 2 ms

^{–2}. [Option (b)]**(3)**Which one among the following acceleration–time graphs most closely represents the motion of the particle?

The acceleration is positive and uniform initially. Afterwards the acceleration becomes zero (since the velocity remains constant) for some time and then becomes negative (since the velocity goes on decreasing) and uniform. The curve shown in (b) therefore represents the motion of the particle.

Let us leave the velocity time graph here and consider a couple of different questions:

**(4)**Two boys running at uniform speeds

*v*

_{1}and

*v*

_{2}respectively along a straight line path in

*opposite*directions get 9 m closer each second. While running along the

*same*direction with their speeds reduced by 50%, they get 0.5m closer each second. The speeds

*v*

_{1}and

*v*

_{2}are respectively

(a) 6 ms

^{–1}**and 3 ms**^{–1}
(b) 5 ms

^{–1}**and 4 ms**^{–1}
(c) 5 ms

^{–1}**and 4.5 ms**^{–1}
(d) 5 ms

^{–1}**and 3.5 ms**^{–1}
(e)

**5.5 ms**^{–1}**and 3.5 ms**^{–1}
This is a simple question involving relative velocity. Wile running along opposite directions, we have

*v*

_{1}+

*v*

_{2}= 9

While running along the same direction, we have

*v*

_{1}/2 –

*v*

_{2}/2 = 0.5, from which

*v*

_{1}–

*v*

_{2}= 1

Solving the above equations, we obtain

*v*_{1}= 5 ms^{–1}**and***v*_{2}= 4 ms^{–1}**(5)**The rear end of a train running on a straight track with uniform acceleration has velocities 6 ms

^{–1}and 10 ms

^{–1}respectively when passing points A and B in its path. The velocity of the rear end midway between these points is approximately

(a) 7 ms

^{–1}
(b) 7.5 ms

^{–1}
(c) 8ms

^{–1}
(d) 8.2 ms

^{–1}
(e) 8.4 ms

^{–1}
We have

*v*^{2}=*u*^{2}+ 2*as*where*u*and*v*are the initial ans final velocities respectively,*a*is the acceleration and*s*is the displacement.
If the distance between A and B is

*s*, we have
10

^{2}= 6^{2}+ 2*as*from which 2*as*= 64
If

*v*_{1}is the velocity midway between A and B we have*v*

_{1}

^{2}= 6

^{2}+ 2

*a*(

*s*/2) = 6

^{2}+ 32 = 68

Therefore,

*v*_{1}= 8.2 ms^{–1}nearly.
You will find more questions (with solution) on one dimensional motion and other sections at AP Physics Resources

Verry nice post. I just stumbled upon your blog and wanted to say tht I

ReplyDeletehave truly enjoyed surfing around your blog posts. After

alll I'll be subscribing to your rss feed and I hope you rite again very soon!

Here is my web site past subjunctive Spanish dar past tense