
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
217.2k+ 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.
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