Average velocity and average speed MCQ Quiz - Objective Question with Answer for Average velocity and average speed - Download Free PDF

The Correct answer is Statement I is incorrect but Statement II is true.

Key Points

Statement I : Area under velocity- time graph gives the distance traveled by the body in a given time.\(\overrightarrow{\mathrm{S}}=\int \overrightarrow{\mathrm{V}} d t\)

Therefore area under the velocity time graph gives displacement.

Hence statement I is false.

Statement II : Area under acceleration- time graph is equal to the change in velocity- in the given time.

a = \(\frac{d V}{d t}\)

⇒ dV = adt

⇒ \(\int d \mathrm{~V}=\int a d t\)

Hence statement - II is correct.

Concept:

• Distance travelled is the product of speed and the time taken while travelling with that speed. 

\( speed = \frac{distance}{time}\)

• The average speed is defined as the total distance travelled by the total time taken during the process. 

\(Average \ speed = \frac{total \ distance}{total \ time}\)

Calculation:

Let the distance from X to Y is d

So, the distance travelled from X to Y or Y to X is d. 

Total distance hence travelled is d + d = 2d

Speed of Car moving from X to Y is V1

distance from  moving X to Y is d

time taken

 \(t_1 = \frac{d}{v_1}\)      ----(1)

Similarly, time taken from moving Y to X is 

\(t_2 = \frac{d}{v_2}\)      ----(2)

total time taken t

\(t = t_1 + t_2 = \frac{d}{v_1} + \frac{d}{v_2}\)

Average speed is the total distance by the total time

\(s= \frac{2d}{ \frac{d}{v_1} + \frac{d}{v_2}}\)

\(\implies s= \frac{2d}{d \ \left ( \frac{1}{v_1} + \frac{1}{v_2} \right )}\)

\(\implies s= \frac{2V_1 V_2}{V_1 + V_2}\)

Clearly, the average speed is the harmonic mean. 

Additional Information

If H is the Harmonic mean of numbers a and b and is given by

 \({\rm{H}} = \frac{{2{\rm{ab}}}}{{{\rm{a}} + {\rm{b}}}}\)

Arithmetic mean = Sum of terms/Number of terms

Geometric mean = G.M. = \(\rm\sqrt{ab}\) 

Concept Used:

The average speed for a round trip, where the speeds are different for the two parts of the journey, is calculated using the formula:

Average speed = (2 × u × v) / (u + v)

Calculation:

We have:

⇒ Average speed = (2 × u × v) / (u + v)

∴ The correct answer is: (2 × u × v) / (u + v)

Concept:

Average speed:

• The total distance travelled divided by the total time taken is called average speed. 

\(Average \ speed = \frac{total \ distance}{total \ time}\)      ----(1)

• The average speed is for the whole Journey. 
• The speed at any instant may change. 
• If the average speed is 35 km/hr for two hours, that not necessarily means 35 km is travelled in one hour. 
• It is possible the speed might have been higher or lower than the 35 km/hr. But, it is the ratio of total distance by total time. 

Calculation:

From equation (1). 

total distance = Average speed × total time. 

If speed = 35 km / hr and time is 2 hr, distance = 35 km/hr × 2 hr = 70 km. 

f time is 3 hr, distance = 35 km/hr × 3 hr = 105 km / hr

So, for first three-hour distance is 105 km and 2 hours it is 70 km. 

Conclusion:

• By calculation, A and D is correct. 
• Statement B says he must have travelled 35 km in first hour is not correct. In first hour the speed may have been different from the average speed of 2 hours. So, it is not correct. 
• Statement C is correct. His speed may change during the journey as average speed is for a certain distance not at any instant, 

So, A, C, D are correct statements. 

CONCEPT:

• Velocity: The rate of change of position i.e. rate of displacement with time is called velocity.

\(v = \frac{{{x_2} - {x_1}}}{{{t_2} - {t_1}}} = \frac{{{\rm{\Delta }}x}}{{{\rm{\Delta }}t}}=\tan{\theta}\)

Where x2 = displacement of the object at t2 and x1 = displacement of the object at t1

• It is a vector quantity having the symbol v.
• A slope of a tangent on the position-time graph represents the velocity of the particle.

EXPLANATION:

• The speeds of the body when it moves from O to A  is

\(\Rightarrow V_{OA}=\frac{AD}{OD}=\frac{2}{1}=2\,m/s\)

• The speeds of the body when it moves from B to C  is

\(\Rightarrow V_{BC}=\frac{CG}{BG}=\frac{2}{0.5}=4\,m/s\)

CONCEPT:

• Average speed: The total path length travelled divided by the total time interval during which the motion has taken place is called the average speed of the particle.

Total distance travelled by the object = 16 m + 16 m = 32 m

Total time taken = 4 s + 2 s = 6 s

\(Average\;speed\;\left( {\bar v} \right) = \frac{{total\;path\;length\;\left( S \right)}}{{total\;time\;taken\;\left( t \right)}}=\frac{32}{6}=5.33\, m/s\)

• Therefore, the average speed of the object is 5.33 m/s.

Formula Used:

Distance = Speed × Time

Calculations:

Let the speed of the train be ‘x’ m/s and the length of the train be ‘y’ m

A train crossed a 120 m long platform in 12 seconds,

⇒ 12 × (x) = 120 + y       ----(1)

A train crossed a 165 m long platform in 15 seconds,

⇒ 15 × (x) = 165 + y       ----(2)

Equation (2) - Equation (1) we get,

⇒ 3x = 45

⇒ x = 15 m/s = 15 × (18/5) km/h = 54 km/h

Concept:

• Distance travelled is the product of speed and the time taken while travelling with that speed. 

\( speed = \frac{distance}{time}\)

• The average speed is defined as the total distance travelled by the total time taken during the process. 

\(Average \ speed = \frac{total \ distance}{total \ time}\)

Calculation:

Given, the distance travelled is s

time interval t = t₂ - t₁ 

Average speed V_av

\(Average \ speed = \frac{total \ distance}{total \ time}\)

So

\(V_{av} = \frac{s}{t_2 - t_1}\)

The correct option is 

\(\frac{s}{t_2 - t_1}\)

Concept:

Equation of motion:

• Equations of motion of kinematics describe the basic concept of the motion of an object such as the position, velocity, or acceleration of an object at various times.
• These three equations of motion govern the motion of an object in 1D, 2D, and 3D.
• Three equations:  v = u + at, s = ut + 0.5 at², 2as = v² - u²
• Here, u = initial velocity, v = final velocity, s = displacement, t = time
• Under the influence of gravity, acceleration a is replaced by acceleration due to gravity g.

Calculation:

Given,

Initial velocity, u = 0 m/s, Acceleration, a = 1.6 m/s²

From the second equation of motion,

s = ut +0.5 at²

12.8 = 0.8 t₁²

\(t_1^2=\frac{12.8}{0.8}=16\)

t₁ = 4 sec.

For s = 20

20 = 0 × t₂ + 0.5 × 1.6 t₂2

t₂ = 5 sec.

The average velocity can be calculated as,

\(v_{avg}=\frac{total\, \, displacement}{total\, \, time\, duration}\)

\(v_{avg}=\frac{20 - 12.8}{5 - 4}=7.2m/s\)

Hence, What is its average velocity 7.2m/s.

CONCEPT:

• Average speed: The total path length travelled divided by the total time interval during which the motion has taken place is called the average speed of the particle.

\(Average\;speed\;\left( {\bar v} \right) = \frac{{total\;path\;length\;\left( S \right)}}{{total\;time\;taken\;\left( t \right)}}\)

​EXPLANATION:

Given -  Initial odometer reading (s₁) = 369 km,  final odometer reading (s₂) = 469 km, and Time (t) = 2 hour

• Mathematically Average speed is written as

\(\Rightarrow v = \frac{\Delta s}{t}=\frac{469-369}{2}=50\,km/hr\)

• To convert the km/hr to m/s, multiply km/hr by 5/18, therefore  

\(\Rightarrow v = 50\times \frac{5}{18}\approx14\, m/s\)

NOTE:

• Odometer or odograph: It is a device used to measure the distance travelled by the vehicle.
• Speedometer or speed meter: A device used by the vehicle to measure the speed of the vehicle.

CONCEPT:

• Average velocity: The average velocity of an object is defined as the displacement per unit of time.
• Let x₁ and x₂ be its positions at instants t₁ and t₂ respectively.
• Then mathematically we can express average velocity (v) as:

\(\Rightarrow v = \frac{displacement}{Time taken}\)

\(\Rightarrow v= \frac{x_{2} - x_{1}}{t_{2} - t_{1}}= \frac{Δ x}{Δ t}\)

where x₂ - x₁, signifies a change in position (denoted by Δx) and t₂ - t₁ is the corresponding change in time (denoted by Δt).

• Average velocity can be represented as V_av also.
• The average speed of an object is obtained by dividing the total distance travelled by the total time taken:

\(\Rightarrow Average \; speed = \frac{Total \; distance \; travelled}{total\; time \; taken}\)

• If the motion is in the same direction along a straight line, the average speed is the same as the magnitude of the average velocity. 

CALCULATION:

Given that:

The total length of the track = 300m. 

Time is taken to cover this length = 200s

\(\Rightarrow Average \; speed= \frac{Total \; distance \; travelled}{total\; time \; taken}\)

\(\Rightarrow Average \; speed= \frac{300}{200} ms^{-1} = 1.5 ms^{-1}\)

• As the person comes back to the same point, the net displacement is zero. Therefore, the average velocity is also zero.
• option 1 is the correct answer.

CONCEPT:

• Distance (S): The total distance traveled by an object from a starting point is called path length. Thus the total path length of an object is called distance traveled by that object.
• Displacement (S’): The minimum path length between the starting point to the final point is called displacement.

• Average speed: The total path length traveled divided by the total time interval during which the motion has taken place is called the average speed of the particle.

\(Average\;speed\;\left( {\bar v} \right) = \frac{{total\;path\;length\;\left( S \right)}}{{total\;time\;taken\;\left( t \right)}}\)

• Average velocity: The ratio of net displacement to total time taken is called average velocity.

\(Average\;velocity\;\left( {\bar V} \right) = \frac{{{\rm{Net\;displacement\;}}\left( {{\rm{S'}}} \right)}}{{time\;taken\;\left( t \right)}}\)

CALCULATION:

• As Mahesh goes to the office and returns back to his house, so the initial point and final point is same for him.

So net displacement of Mahesh (S) = 0

Total time taken (t) = 10 + 15 = 25 min

Average velocity = (Net displacement)/(total time taken) = 0/25 = 0

Hence option 4 is correct.

Concept:

• Displacement (x): The change in the position of an object is called displacement.
  • Displacement is a one-dimensional quantity representing the smallest separation between two defined points.
  • The standard unit of displacement in the International System of Units (SI) is the meter (m).

• Velocity (V): The rate of change of displacement is called velocity. It is a vector quantity. 
  • The SI unit of velocity is called m/s and the CGS unit is cm/s.
  • The slope of the displacement-time graph gives the velocity

\(Instanteneous\;Velocity\;\left( v \right) = \frac{{dx}}{{dt}}\)

Hence, dx = v⋅dt

\(Displacement\;\left( X \right) = \mathop \smallint \nolimits_{{x_1}}^{{x_2}} dx = \mathop \smallint \nolimits_{{t_1}}^{{t_2}} V \cdot dt\)

• Thus the area under the velocity-time graph gives the displacement. Similarly, the area under-speed time graph gives the total distance travelled

Calculation:

Area under the velocity-time curve represents displacement.

To get exact position at t, we need to calculate the area of the shaded part in the curve as shown below

Area of trapezium = ½ × (Sum of parallel sides) × distance between them.

⇒ Area of trapezium = ½ × (15 + 5) × 100

⇒ Area of trapezium = ½ × 20 × 100 = 1000 m

Hence the correct option is 1000 meter.

Concept: 

• Speed: The distance traveled by the body in unit time.

\(Speed = \frac{{Distance}}{{Time}}\)

Also, \(Time= \frac{{Distance}}{{Speed}}\)

Average speed: 

• The total path length traveled divided by the total time interval during which the motion has taken place is called the average speed of the particle.

\(\left( {\bar v} \right) = \frac{{total\;path\;length\;\left( S \right)}}{{total\;time\;taken\;\left( t \right)}}\)

Calculation:

V_max = 120 km/hr

Distance = 372 km

Time (t) = 7 hr 45 min

= 7+ (45/60) =  (31/4) Hr

\(Average\;Speed = \frac{{Total\;Distance}}{{Total\;Time}}\) = \( \frac{{372\times4}}{{31}}\)

V_avg = 48 (km/hr)

Hence the required difference is  = 120 - 48 = 72 km/hr

CONCEPT:

• Speed: The distance travelled by the body in unit time.
• \(Speed = \frac{{Distance}}{{Time}}\)
• Also, \(Time= \frac{{Distance}}{{Speed}}\)
• Average Speed: Since the body does not always have the same speed, So we use the concept of average speed. It is the ratio of total distance covered to total time taken by the body.
• \(Average\;Speed = \frac{{Total\;Distance}}{{Total\;Time}}\)

CALCULATION:

• Total Distance = 4 km
• Distance travelled in first 10 min at speed of 12 km/hr = Speed × time = 12 × (10/60) = 2 km
• Time taken to cover the distance of last 2 km = Distance/speed = 2/8 = 1/4 = 0.25 hr = 15 min
• Total time = 10 min + 15 min (Chat with friend) + 15 min = 40 min = 2/3 hr

We know that,

\(Average\;Speed = \frac{{Total\;Distance}}{{Total\;Time}}\)

\(⇒ Average\;Speed = \frac{{4\;km}}{{2/3\;hr}} = 6 \;km/hr\)

⇒ Average Speed = 6 km/hr