
A bus starts from rest moving with an acceleration of \[2{\rm{ m/}}{{\rm{s}}^2}\]. A cyclist, 96 m behind the bus, starts simultaneously towards the bus at \[20{\rm{ m/s}}\]. After what time will he be able to overtake the bus:-
A. 8 s
B. 10 s
C. 12 s
D. 1 s
Answer
162.9k+ views
Hint:The second equation of motion gives the total distance covered by an object which is moving with uniform acceleration. Here the bus and cyclist both are in motion so the concept of relative velocity is used. The relative velocity is defined as the velocity of an object with respect to another object. Also, we say that It is the time rate of change of relative position of one object with respect to another object.
Formula used Relative velocity of c with respect to b is given as,
\[{v_{cb}} = {v_c} - {v_b}\]
The second equation of motion is given as:
\[s = ut + \dfrac{1}{2}a{t^2}\]
Where s is the distance travelled, u is the initial velocity, t is the time taken and a is acceleration.
Complete step by step solution:
Velocity of bus, \[{v_b} = 0\]
Velocity of bus, \[{v_c} = 20\,m/s\]
Acceleration of bus,\[{a_b} = 2\,m/{s^2}\]
Acceleration of bus,\[{a_c} = 0\,m/{s^2}\]
Relative separation, \[s = 96\,m\]
Relative velocity is given as,
\[{v_{cb}} = {v_c} - {v_b}\]
\[\Rightarrow {v_{cb}} = 20 - 0\]
\[\Rightarrow {v_{cb}} = 20\,m/s\]
Relative acceleration is given as,
\[{a_{cb}} = {a_c} - {a_b}\]
\[\Rightarrow {a_{cb}} = 0 - 2\]
\[\Rightarrow {a_{cb}}= - 2\,m/s\]
By using the second equation of motion,
\[s = ut + \dfrac{1}{2}a{t^2}\]
Substituting the values
\[96 = 20t + \dfrac{1}{2} \times 2 \times {t^2}\]
\[\Rightarrow {t^2} - 20t + 96 = 0\]
By solving the equation, we get
\[t = 8s\] and \[t = 12s\]
Now here we have 8s which is the minimum value of time. Because a cyclist wants to overtake the bus. So we take the minimum value of time which is 8s. Therefore, after 8 second time, a cyclist will be able to overtake the bus
Hence option A is the correct answer.
Note: To determine the complete motion of an object, we have to consider the effect on the object due to the medium. We can calculate the relative velocity of the object by considering the velocity of the particle and the velocity of the medium.
Formula used Relative velocity of c with respect to b is given as,
\[{v_{cb}} = {v_c} - {v_b}\]
The second equation of motion is given as:
\[s = ut + \dfrac{1}{2}a{t^2}\]
Where s is the distance travelled, u is the initial velocity, t is the time taken and a is acceleration.
Complete step by step solution:
Velocity of bus, \[{v_b} = 0\]
Velocity of bus, \[{v_c} = 20\,m/s\]
Acceleration of bus,\[{a_b} = 2\,m/{s^2}\]
Acceleration of bus,\[{a_c} = 0\,m/{s^2}\]
Relative separation, \[s = 96\,m\]
Relative velocity is given as,
\[{v_{cb}} = {v_c} - {v_b}\]
\[\Rightarrow {v_{cb}} = 20 - 0\]
\[\Rightarrow {v_{cb}} = 20\,m/s\]
Relative acceleration is given as,
\[{a_{cb}} = {a_c} - {a_b}\]
\[\Rightarrow {a_{cb}} = 0 - 2\]
\[\Rightarrow {a_{cb}}= - 2\,m/s\]
By using the second equation of motion,
\[s = ut + \dfrac{1}{2}a{t^2}\]
Substituting the values
\[96 = 20t + \dfrac{1}{2} \times 2 \times {t^2}\]
\[\Rightarrow {t^2} - 20t + 96 = 0\]
By solving the equation, we get
\[t = 8s\] and \[t = 12s\]
Now here we have 8s which is the minimum value of time. Because a cyclist wants to overtake the bus. So we take the minimum value of time which is 8s. Therefore, after 8 second time, a cyclist will be able to overtake the bus
Hence option A is the correct answer.
Note: To determine the complete motion of an object, we have to consider the effect on the object due to the medium. We can calculate the relative velocity of the object by considering the velocity of the particle and the velocity of the medium.
Recently Updated Pages
JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Charging and Discharging of Capacitor

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line
