Speed and Velocity
Category : NEET
Speed and Velocity
(1) Speed: Rate of distance covered with time is called speed.
(i) It is a scalar quantity having symbol \[\upsilon \].
(ii) Dimension: \[\left[ {{M}^{0}}{{L}^{1}}{{T}^{1}} \right]\]
(iii) Unit: metre/second (S.I.), cm/second (C.G.S.)
(iv) Types of speed:
(a) Uniform speed: When a particle covers equal distances in equal intervals of time, (no matter how small the intervals are) then it is said to be moving with uniform speed. In given illustration motorcyclist travels equal distance (= 5 m) in each second. So we can say that particle is moving with uniform speed of 5 m/s.
(b) Non-uniform (variable) speed: In non-uniform speed particle covers unequal distances in equal intervals of time. In the given illustration motorcyclist travels 5m in 1st second, 8m in 2nd second, 10m in 3rd second, 4m in 4th second etc.
Therefore its speed is different for every time interval of one second. This means particle is moving with variable speed.
(c) Average speed: The average speed of a particle for a given ‘Interval of time’ is defined as the ratio of distance travelled to the time taken.
\[Average\text{ }speed=\frac{\text{Distance travelled}}{\text{Time taken}}\]; \[{{v}_{av}}=\frac{\Delta s}{\Delta t}\]
q Time average speed: When particle moves with different uniform speed \[{{\upsilon }_{1}},\,\,{{\upsilon }_{2}},\,\,{{\upsilon }_{3}}\]... etc. in different time intervals\[{{t}_{1}},\,\,{{t}_{2}},\,\,{{t}_{3}}\], ... etc. respectively, its average speed over the total time of journey is given as
\[{{v}_{av}}=\frac{\text{Total distance covered}}{\text{Total time elapsed}}=\frac{{{d}_{1}}+{{d}_{2}}+{{d}_{3}}+......}{{{t}_{1}}+{{t}_{2}}+{{t}_{3}}+......}=\frac{{{\upsilon }_{1}}{{t}_{1}}+{{\upsilon }_{2}}{{t}_{2}}+{{\upsilon }_{3}}{{t}_{3}}+......}{{{t}_{1}}+{{t}_{2}}+{{t}_{3}}+......}\]
Special case: When particle moves with speed \[{{v}_{1}}\] upto half time of its total motion and in rest time it is moving with speed \[{{v}_{2}}\] then \[{{v}_{av}}=\frac{{{v}_{1}}+{{v}_{2}}}{2}\]
q Distance averaged speed: When a particle describes different distances \[{{d}_{1}},\,\,{{d}_{2}},\,\,{{d}_{3}}\], ...... with different time intervals \[{{t}_{1}},\,\,{{t}_{2}},\,\,{{t}_{3}}\], ...... with speeds \[{{v}_{1}},{{v}_{2}},{{v}_{3}}......\] respectively then the speed of particle averaged over the total distance can be given as
\[{{\upsilon }_{av}}=\frac{\text{Total distance covered}}{\text{Total time elapsed}}=\frac{{{d}_{1}}+{{d}_{2}}+{{d}_{3}}+......}{{{t}_{1}}+{{t}_{2}}+{{t}_{3}}+......}=\frac{{{d}_{1}}+{{d}_{2}}+{{d}_{3}}+......}{\frac{{{d}_{1}}}{{{\upsilon }_{1}}}+\frac{{{d}_{2}}}{{{\upsilon }_{2}}}+\frac{{{d}_{3}}}{{{\upsilon }_{3}}}+......}\]
q When particle moves the first half of a distance at a speed of v1 and second half of the distance at speed v2 then
\[{{v}_{av}}=\frac{2{{v}_{1}}{{v}_{2}}}{{{v}_{1}}+{{v}_{2}}}\]
q When particle covers one-third distance at speed v1, next one third at speed v2 and last one third at speed v3, then
\[{{v}_{av}}=\frac{3\,{{v}_{1}}{{v}_{2}}{{v}_{3}}}{{{v}_{1}}{{v}_{2}}+{{v}_{2}}{{v}_{3}}+{{v}_{3}}{{v}_{1}}}\]
(d) Instantaneous speed: It is the speed of a particle at particular instant. When we say “speed”, it usually means instantaneous speed.
The instantaneous speed is average speed for infinitesimally small time interval (i.e., \[\Delta t\to 0\]). Thus
Instantaneous speed \[v=\underset{\Delta t\to 0}{\mathop{\lim }}\,\,\,\frac{\Delta s}{\Delta t}=\frac{ds}{dt}\]
(2) Velocity: Rate of change of position i.e. rate of displacement with time is called velocity.
(i) It is a scalar quantity having symbol \[v\].
(ii) Dimension: \[\left[ {{M}^{0}}{{L}^{1}}{{T}^{1}} \right]\]
(iii) Unit: metre/second (S.I.), cm/second (C.G.S.)
(iv) Types
(a) Uniform velocity: A particle is said to have uniform velocity, if magnitudes as well as direction of its velocity remains same and this is possible only when the particles moves in same straight line without reversing its direction.
(b) Non-uniform velocity: A particle is said to have non-uniform velocity, if either of magnitude or direction of velocity changes (or both changes).
(c) Average velocity: It is defined as the ratio of displacement to time taken by the body
\[\text{Average velocity}=\frac{\text{Displacement}}{\text{Time taken}};\,\,\,\,\,\,{{\vec{v}}_{av}}=\frac{\Delta \vec{r}}{\Delta t}\]
(d) Instantaneous velocity: Instantaneous velocity is defined as rate of change of position vector of particles with time at a certain instant of time.
Instantaneous velocity \[\vec{v}=\underset{t\to 0}{\mathop{\lim }}\,\,\,\frac{\Delta \,\vec{r}}{\Delta t}=\frac{d\vec{r}}{dt}\]
(v) Comparison between instantaneous speed and instantaneous velocity
(a) Instantaneous velocity is always tangential to the path followed by the particle.
When a stone is thrown from point O then at point of projection the instantaneous velocity of stone is \[{{v}_{1}}\] , at point A the instantaneous velocity of stone is \[{{v}_{2}}\], similarly at point B and C are \[{{v}_{3}}\,\,and\,\,{{v}_{4}}\] respectively.
Direction of these velocities can be found out by drawing a tangent on the trajectory at a given point.
(b) A particle may have constant instantaneous speed but variable instantaneous velocity.
Example: When a particle is performing uniform circular motion then for every instant of its circular motion its speed remains constant but velocity changes at every instant.
(c) The magnitude of instantaneous velocity is equal to the instantaneous speed.
(d) If a particle is moving with constant velocity then its average velocity and instantaneous velocity are always equal.
(e) If displacement is given as a function of time, then time derivative of displacement will give velocity.
Let displacement \[\vec{x}={{A}_{0}}-{{A}_{1}}t+{{A}_{2}}{{t}^{2}}\]
Instantaneous velocity \[\vec{v}=\frac{d\vec{x}}{dt}=\frac{d}{dt}\,({{A}_{0}}-{{A}_{1}}t+{{A}_{2}}{{t}^{2}})\]
\[\vec{v}=-{{A}_{1}}+2{{A}_{2}}t\]
For the given value of t, we can find out the instantaneous velocity.
e.g. for \[t=0\], Instantaneous velocity \[\vec{v}=-{{A}_{1}}\] and Instantaneous speed \[|\vec{v}|\,={{A}_{1}}\]
(vi) Comparison between average speed and average velocity
(a) Average speed is scalar while average velocity is a vector both having same units (m/s) and dimensions\[[L{{T}^{-1}}]\].
(b) Average speed or velocity depends on time interval over which it is defined.
(c) For a given time interval average velocity is single valued while average speed can have many values depending on path followed.
(d) If after motion body comes back to its initial position then \[{{\vec{v}}_{av}}=\vec{0}\] (as \[\Delta \vec{r}=0\]) but \[{{v}_{av}}>\vec{0}\] and finite as\[(\Delta s>0)\].
(e) For a moving body average speed can never be negative or zero (unless \[t\to \infty )\] while average velocity can be i.e. \[{{v}_{av}}>0\] while \[{{\vec{v}}_{a\upsilon }}= or < 0.\]
Sample problems based on speed and velocity
Problem 4. If a car covers 2/5th of the total distance with v1 speed and 3/5th distance with \[{{v}_{2}}\] then average speed is [MP PMT 2003]
(a) \[\frac{1}{2}\sqrt{{{v}_{1}}{{v}_{2}}}\] (b) \[\frac{{{v}_{1}}+{{v}_{2}}}{2}\] (c) \[\frac{2{{v}_{1}}{{v}_{2}}}{{{v}_{1}}+{{v}_{2}}}\] (d) \[\frac{5{{v}_{1}}{{v}_{2}}}{3{{v}_{1}}+2{{v}_{2}}}\]
Solution: (d) \[Average\text{ }speed=\,\frac{\text{Total distance travelled}}{\text{Total time taken}}=\frac{x}{{{t}_{1}}+{{t}_{2}}}\]
\[=\,\,\frac{x}{\frac{(2/5)\,x}{{{v}_{1}}}+\frac{(3/5)x}{{{v}_{2}}}}\,=\frac{5{{v}_{1}}{{v}_{2}}}{2{{v}_{2}}+3{{v}_{1}}}\]
Problem 5. A car accelerated from initial position and then returned at initial point, then [AIEEE 2002]
(a) Velocity is zero but speed increases (b) Speed is zero but velocity increases
(c) Both speed and velocity increase (d) Both speed and velocity decrease
Solution: (a) As the net displacement = 0
Hence velocity = 0; but speed increases.
Note: \[\frac{|\text{Average velocity}|}{|\text{Average }\,\text{speed}|}\le 1\Rightarrow \,\,|\text{Av}\text{.}\,\text{speed}|\,\,\ge \,\,|\text{Av}\text{.}\,\text{velocity}|\]
Problem 6. A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km/h. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km/h. The average speed of the man over the interval of time 0 to 40 min. is equal to [AMU (Med.) 2002]
(a) 5 km/h (b) \[\frac{25}{4}\,km/h\] (c) \[\frac{30}{4}km/h\] (d) \[\frac{45}{8}\,km/h\]
Solution: (d) Time taken in going to market \[=\frac{2.5}{5}=\frac{1}{2}hr=30\min .\]
As we are told to find average speed for the interval 40 min., so remaining time for consideration of motion is 10 min.
So distance travelled in remaining 10 min \[=\,\,7.5\times \frac{10}{60}=1.25\,\,km.\]
Hence, average speed \[=\frac{\text{Total distance }}{\text{Total time}}=\frac{(2.5+1.25)\,km}{(40/60)\,hr.}=\frac{45}{8}km/hr\]
Problem 7. The relation \[3t=\sqrt{3x}+6\] describes the displacement of a particle in one direction where \[x\] is in metres and t in sec. The displacement, when velocity is zero, is [CPMT 2000]
(a) 24 metres (b) 12 metres (c) 5 metres (d) Zero
Solution: (d) \[3t=\sqrt{3x}+6\Rightarrow \sqrt{3x}=(3t-6)\Rightarrow 3x={{(3t-6)}^{2}}\Rightarrow x=3{{t}^{2}}-12t+12\]
\[\therefore \,\,\,\,v=\frac{dx}{dt}=\frac{d}{dt}(3{{t}^{2}}-12t+12)=6t-12\]
If velocity = 0 then, \[6t-12=0\] \[\Rightarrow \,t=2sec\]
Hence at t = 2, x = 3(2)2 – 12 (2) + 12 = 0 metres.
Problem 8. The motion of a particle is described by the equation \[x=a+b{{t}^{2}}\] where \[a=15\]cm and \[b=3\]cm. Its instantaneous velocity at time 3 sec will be [AMU (Med.) 2000]
(a) 36 cm/sec (b) 18 cm/sec (c) 16 cm/sec (d) 32 cm/sec
Solution: (b) \[x=a+b{{t}^{2}}\] \ v =\[\frac{dx}{dt}=0+2bt\]
At t = 3 sec, v = \[2\times 3\times 3\] = \[18cm/sec\] (As \[b=3cm\])
Problem 9. A train has a speed of 60 km/h for the first one hour and 40 km/h for the next half hour. Its average speed in km/h is [JIPMER 1999]
(a) 50 (b) 53.33 (c) 48 (d) 70
Solution: (b) Total distance travelled = \[60\times 1+40\times \frac{1}{2}=80\,km\] and Total time taken = \[1\,hr+\frac{1}{2}hr=\frac{3}{2}hr\]
\[\therefore \,\,\,\,Average\text{ }speed=\frac{80}{{3}/{2}\;}=53.33\,\,km/h\]
Problem 10. A person completes half of its his journey with speed \[{{\upsilon }_{1}}\] and rest half with speed \[{{\upsilon }_{2}}\]. The average speed of the person is [RPET 1993; MP PMT 2001]
(a) \[\upsilon =\frac{{{\upsilon }_{1}}+{{\upsilon }_{2}}}{2}\] (b) \[\upsilon =\frac{2{{\upsilon }_{1}}\,{{\upsilon }_{2}}}{{{\upsilon }_{1}}+{{\upsilon }_{2}}}\] (c) \[\upsilon =\frac{{{\upsilon }_{1}}\,{{\upsilon }_{2}}}{{{\upsilon }_{1}}+{{\upsilon }_{2}}}\] (d) \[\upsilon =\sqrt{{{\upsilon }_{1}}\,{{\upsilon }_{2}}}\]
Solution: (b) In this problem total distance is divided into two equal parts. So
\[{{\upsilon }_{av}}=\frac{{{d}_{1}}+{{d}_{2}}}{\frac{{{d}_{1}}}{{{\upsilon }_{1}}}+\frac{{{d}_{2}}}{{{\upsilon }_{2}}}}=\frac{\frac{d}{2}+\frac{d}{2}}{\frac{d/2}{{{\upsilon }_{1}}}+\frac{d/2}{{{\upsilon }_{2}}}}\Rightarrow {{\upsilon }_{av}}=\frac{2}{\frac{1}{{{\upsilon }_{1}}}+\frac{1}{{{\upsilon }_{2}}}}=\frac{2{{\upsilon }_{1}}\,{{\upsilon }_{2}}}{{{\upsilon }_{1}}+{{\upsilon }_{2}}}\]
Problem 11. A car moving on a straight road covers one third of the distance with 20 km/hr and the rest with 60 km/hr. The average speed is [MP PMT 1999; CPMT 2002]
(a) 40 km/hr (b) 80 km/hr (c) \[46\frac{2}{3}km/hr\] (d) 36 km/hr
Solution: (d) Let total distance travelled = x and total time taken \[{{t}_{1}}+\text{ }{{t}_{2}}=\frac{x/3}{20}+\frac{2x/3}{60}\]
\[\therefore \,\,\,\,Average\text{ }speed=\frac{x}{\frac{(1/3)x}{20}+\frac{(2/3)x}{60}}=\frac{1}{\frac{1}{60}+\frac{2}{180}}=36\,km/hr\]
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