Equations of Kinematics
Category : NEET
Equations of Kinematics
These are the various relations between u, v, a, t and s for the moving particle where the notations are used as:
u = Initial velocity of the particle at time t = 0 sec
v = Final velocity at time t sec
a = Acceleration of the particle
s = Distance travelled in time t sec
sn = Distance travelled by the body in nth sec
(1) When particle moves with zero acceleration
(i) It is a unidirectional motion with constant speed.
(ii) Magnitude of displacement is always equal to the distance travelled.
(iii) v = u, s = u t [As a = 0]
(2) When particle moves with constant acceleration
(i) Acceleration is said to be constant when both the magnitude and direction of acceleration remain constant.
(ii) There will be one dimensional motion if initial velocity and acceleration are parallel or anti-parallel to each other.
(iii) Equations of motion in scalar from Equation of motion in vector from
\[\upsilon =u+at\] \[\vec{v}=\vec{u}+\vec{a}t\]
\[s=ut+\frac{1}{2}a{{t}^{2}}\] \[\vec{s}=\vec{u}t+\frac{1}{2}\vec{a}{{t}^{2}}\]
\[{{\upsilon }^{2}}={{u}^{2}}+2as\] \[\vec{v}.\vec{v}-\vec{u}.\vec{u}=2\vec{a}.\vec{s}\]
\[s=\left( \frac{u+v}{2} \right)\,t\] \[\vec{s}=\frac{1}{2}(\vec{u}+\vec{v})\,t\]
\[{{s}_{n}}=u+\frac{a}{2}\,(2n-1)\] \[{{\vec{s}}_{n}}=\vec{u}+\frac{{\vec{a}}}{2}\,(2n-1)\]
(3) Important points for uniformly accelerated motion
(i) If a body starts from rest and moves with uniform acceleration then distance covered by the body in t sec is proportional to t2 (i.e. \[s\propto {{t}^{2}}\]).
So we can say that the ratio of distance covered in 1 sec, 2 sec and 3 sec is \[{{1}^{2}}:{{2}^{2}}:{{3}^{2}}\] or 1 : 4 : 9.
(ii) If a body starts from rest and moves with uniform acceleration then distance covered by the body in nth sec is proportional to \[(2n-1)\] (i.e. \[{{s}_{n}}\propto \,(2n-1)\]
So we can say that the ratio of distance covered in I sec, II sec and III sec is 1 : 3 : 5.
(iii) A body moving with a velocity u is stopped by application of brakes after covering a distance s. If the same body moves with velocity nu and same braking force is applied on it then it will come to rest after covering a distance of \[{{\operatorname{n}}^{2}}s\].
As \[{{\upsilon }^{2}}={{u}^{2}}-2as\Rightarrow 0={{u}^{2}}-2as\Rightarrow s=\frac{{{u}^{2}}}{2a},\,\,\,s\propto {{u}^{2}}\] [since a is constant]
So we can say that if u becomes n times then s becomes \[{{n}^{2}}\] times that of previous value.
(iv) A particle moving with uniform acceleration from A to B along a straight line has velocities \[{{\upsilon }_{1}}\,\,and\,\,\,{{\upsilon }_{2}}\] at A and B respectively. If C is the mid-point between A and B then velocity of the particle at C is equal to
\[\upsilon =\sqrt{\frac{\upsilon _{1}^{2}+\upsilon _{2}^{2}}{2}}\]
Sample problems based on uniform acceleration
Problem 40. A body A moves with a uniform acceleration \[a\] and zero initial velocity. Another body B, starts from the same point moves in the same direction with a constant velocity \[v\]. The two bodies meet after a time t. The value of \[t\] is [MP PET 2003]
(a) \[\frac{2v}{a}\] (b) \[\frac{v}{a}\] (c) \[\frac{v}{2a}\] (d) \[\sqrt{\frac{v}{2a}}\]
Solution: (a) Let they meet after time ‘t’. Distance covered by body \[A=\frac{1}{2}a{{t}^{2}}\]; Distance covered by body \[B=vt\]
and \[\frac{1}{2}a{{t}^{2}}=vt\] \[\therefore \] \[t=\frac{2v}{a}\].
Problem 41. A student is standing at a distance of 50metres from the bus. As soon as the bus starts its motion with an acceleration of 1ms–2, the student starts running towards the bus with a uniform velocity \[u\]. Assuming the motion to be along a straight road, the minimum value of \[u\], so that the students is able to catch the bus is [KCET 2003]
(a) \[5 m{{s}^{1}}\] (b) \[8\text{ }m{{s}^{1}}\] (c) \[10\text{ }m{{s}^{1}}\] (d) \[12\text{ }m{{s}^{1}}\]
Solution: (c) Let student will catch the bus after \[t\]sec. So it will cover distance ut.
Similarly distance travelled by the bus will be \[\frac{1}{2}a{{t}^{2}}\] for the given condition
\[ut=50+\frac{1}{2}a{{t}^{2}}=50+\frac{{{t}^{2}}}{2}\Rightarrow u=\frac{50}{t}\,+\,\frac{t}{2}\] (As \[a=1\,m/{{s}^{2}}\])
To find the minimum value of u, \[\frac{du}{dt}=0\], so we get t = 10 sec
then u = 10 m/s.
Problem 42. A car, moving with a speed of 50 km/hr, can be stopped by brakes after at least 6m. If the same car is moving at a speed of 100 km/hr, the minimum stopping distance is [AIEEE 2003]
(a) 6 m (b) 12 m (c) 18 m (d) 24 m
Solution: (d) \[{{v}^{2}}={{u}^{2}}-2as\Rightarrow 0={{u}^{2}}-2as\Rightarrow s=\frac{{{u}^{2}}}{2a}\Rightarrow s\propto {{u}^{2}}\] (As a = constant)
\[\frac{{{s}_{2}}}{{{s}_{1}}}={{\left( \frac{{{u}_{2}}}{{{u}_{1}}} \right)}^{2}}={{\left( \frac{100}{50} \right)}^{2}}\Rightarrow {{s}_{2}}=4{{s}_{1}}=4\times 12=24\,m.\]
Problem 43. The velocity of a bullet is reduced from 200m/s to 100m/s while travelling through a wooden block of thickness 10cm. The retardation, assuming it to be uniform, will be [AIIMS 2001]
(a) \[10\,\,\times \,\,1{{0}^{4}}m/{{s}^{2}}\] (b) \[12\times {{10}^{4}}m/{{s}^{2}}\] (c) \[13.5\times {{10}^{4}}m/{{s}^{2}}\] (d) \[15\times {{10}^{4}}m/{{s}^{2}}\]
Solution: (d) \[u=200\,m/s\], \[v\,=\,100\,m/s\], \[s=\,0.1\,m\]
\[a=\frac{{{u}^{2}}-{{v}^{2}}}{2s}\,=\,\frac{{{(200)}^{2}}\,-\,{{(100)}^{2}}}{2\times 0.1}\,=\,15\times {{10}^{4}}\,m/{{s}^{2}}\]
Problem 44. A body A starts from rest with an acceleration \[{{a}_{1}}\]. After 2 seconds, another body B starts from rest with an acceleration \[{{a}_{2}}\]. If they travel equal distances in the 5th second, after the start of A, then the ratio \[{{a}_{1}}:{{a}_{2}}\] is equal to [AIIMS 2001]
(a) 5 : 9 (b) 5 : 7 (c) 9 : 5 (d) 9 : 7
Solution: (a) By using \[{{S}_{n}}=u+\frac{a}{2}\left( 2n-1 \right)\], Distance travelled by body A in 5th second = \[0+\frac{{{a}_{1}}}{2}(2\times 5-1)\]
Distance travelled by body B in 3rd second is \[=\,0\,+\frac{{{a}_{2}}}{2}\,(2\times 3-1)\]
According to problem: \[0+\frac{{{a}_{1}}}{2}(2\times 5-1)=\,0\,+\frac{{{a}_{2}}}{2}\,(2\times 3-1)\Rightarrow 9{{a}_{1}}=\,5{{a}_{2}}\Rightarrow \frac{{{a}_{1}}}{{{a}_{2}}}=\frac{5}{9}\]
Problem 45. The average velocity of a body moving with uniform acceleration travelling a distance of 3.06 m is \[0.34 m{{s}^{1}}\]. If the change in velocity of the body is \[0.18\,m{{s}^{1}}\] during this time, its uniform acceleration is [EAMCET (Med.) 2000]
(a) \[0.01 m{{s}^{2}}\] (b) \[0.02 m{{s}^{2}}\] (c) \[0.03 m{{s}^{2}}\] (d) \[0.04 m{{s}^{2}}\]
Solution: (b) \[\text{Time}=\frac{\text{Distance}}{\text{Average velocity}}=\frac{3.06}{0.34}=9\,sec\]
and \[\text{Acceleration}=\frac{\text{Change in velocity}}{\text{Time}}=\,\frac{0.18}{9}\,=\,0.02\,m/{{s}^{2}}\]
Problem 46. A particle travels 10m in first 5 sec and 10m in next 3 sec. Assuming constant acceleration what is the distance travelled in next 2 sec [RPET 2000]
(a) 8.3 m (b) 9.3 m (c) 10.3 m (d) None of above
Solution: (a) Let initial \[(t=0)\] velocity of particle = \[u\]
for first 5 sec of motion \[{{s}_{5}}=10\,metre\], so by using \[s=ut+\frac{1}{2}a\,{{t}^{2}}\]
\[10=5u+\frac{1}{2}a{{(5)}^{2}}\Rightarrow 2u+5a=4\] …. (i)
for first 8 sec of motion \[{{s}_{8}}=20\,metre\]
\[20=8u+\frac{1}{2}a{{(8)}^{2}}\Rightarrow 2u+8a=5\] .... (ii)
By solving (i) and (ii) \[u=\frac{7}{6}m/s\,\,\,a=\frac{1}{3}\,m/{{s}^{2}}\]
Now distance travelled by particle in total 10 sec. \[{{s}_{10}}=u\times 10+\frac{1}{2}a{{\left( 10 \right)}^{2}}\]
by substituting the value of \[u\] and \[a\] we will get \[{{s}_{10}}=28.3\,m\]
So the distance in last 2 sec \[=\text{ }{{s}_{10}}\text{ }{{s}_{8}}=28.3-20=8.3\,m\]
Problem 47. A body travels for 15 sec starting from rest with constant acceleration. If it travels distances \[{{S}_{1}},\ {{S}_{2}}\] and \[{{S}_{3}}\] in the first five seconds, second five seconds and next five seconds respectively the relation between \[{{S}_{1}},\ {{S}_{2}}\] and \[{{S}_{3}}\] is [AMU (Engg.) 2000]
(a) \[{{S}_{1}}={{S}_{2}}={{S}_{3}}\] (b) \[5{{S}_{1}}=3{{S}_{2}}={{S}_{3}}\] (c) \[{{S}_{1}}=\frac{1}{3}{{S}_{2}}=\frac{1}{5}{{S}_{3}}\] (d) \[{{S}_{1}}=\frac{1}{5}{{S}_{2}}=\frac{1}{3}{{S}_{3}}\]
Solution: (c) Since the body starts from rest. Therefore \[u=0\].
\[{{S}_{1}}=\frac{1}{2}a{{(5)}^{2}}=\frac{25a}{2}\]
\[{{S}_{1}}+{{S}_{2}}=\frac{1}{2}a{{(10)}^{2}}=\frac{100a}{2}\Rightarrow {{S}_{2}}=\frac{100a}{2}-{{S}_{1}}=75\frac{a}{2}\]
\[{{S}_{1}}+{{S}_{2}}+{{S}_{3}}=\frac{1}{2}a{{(15)}^{2}}=\frac{225a}{2}\Rightarrow {{S}_{3}}=\frac{225a}{2}-{{S}_{2}}-{{S}_{1}}=\frac{125a}{2}\]
Thus Clearly \[{{S}_{1}}=\frac{1}{3}{{S}_{2}}=\frac{1}{5}{{S}_{3}}\]
Problem 48. If a body having initial velocity zero is moving with uniform acceleration \[8\,m/{{\sec }^{2}},\]the distance travelled by it in fifth second will be [MP PMT 1996; DPMT 2000]
(a) 36 metres (b) 40 metres (c) 100 metres (d) Zero
Solution: (a) \[{{S}_{n}}=u+\frac{1}{2}a(2n-1)=0+\frac{1}{2}(8)\,\,[2\times 5-1]=36\,\,metres\]
Problem 49. The engine of a car produces acceleration 4m/sec2 in the car, if this car pulls another car of same mass, what will be the acceleration produced [RPET 1996]
(a) \[8 m/{{s}^{2}}\] (b) \[2 m/{{s}^{2}}\] (c) \[4 m/{{s}^{2}}\] (d) \[\frac{1}{2}m/{{s}^{2}}\]
Solution: (b) F = ma \[a\propto \,\frac{1}{m}\] if F = constant. Since the force is same and the effective mass of system becomes double
\[\frac{{{a}_{2}}}{{{a}_{1}}}=\frac{{{m}_{1}}}{{{m}_{2}}}=\frac{m}{2m},\,\,{{a}_{2}}=\frac{{{a}_{1}}}{2}=2\,\text{m/}{{\text{s}}^{\text{2}}}\]
Problem 50. A body starts from rest. What is the ratio of the distance travelled by the body during the 4th and 3rd second. [CBSE PMT 1993]
(a) 7/5 (b) 5/7 (c) 7/3 (d) 3/7
Solution: (a) As \[{{S}_{n}}\,\propto \,(2n-1)\], \[\frac{{{S}_{4}}}{{{S}_{3}}}=\frac{7}{5}\]
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