Question

A bus starts from rest moving with an acceleration of 2m/s².A cyclist 96m behind the bus starts simultaneously towards the bus at 20m/s, after what time will be able to overtake the bus?(answer with explanation)... incorrect answer will be reported ☑️☑️✔️​

Answer 1

Given:-

→Initial velocity of the bus = 0

(as it starts from rest)

→Acceleration of the bus = 2m/s²

→Separation between the bus and the

cyclist = 96m

→ Speed of the cyclist = 20m/s

To find:-

→The time after which, the cyclist will be able to overtake the bus.

Solution:-

Let the time after which, the cyclist overtakes the bus be 't'.

Firstly, let's calculate the distance travelled by the bus in time t by using 2nd equation if motion.

=>s = ut+1/2at²

=>s₁= 0(t)+1/2×2×t²

=>s₁=t² ------(1)

Now,let's calculate the distance travelled by the cyclist in time t.

=>Distance=Speed×Time

=>s₂ = 20×t

=>s₂ = 20t ------(2)

Since,the cyclist was 96m behind the bus.Thus:-

=>s₁ = s₂-96

=>t² = 20t-96 [∵s₁ = t² , s₂ =20t]

=>t²-20t+96 = 0

=>t²-12t-8t+96 = 0

=>(t-12)(t-8)=0

∴ t=12s or t=8s

We can see here,that the bus and the cyclist are meeting each other at 2 instances of time.

Hence,we can infer that at t=8s,the cyclist overtakes the bus and at t=12s,the bus again overtakes the cyclist.

Thus,the cyclist will be able to overtake the bus after 8 seconds.

Answer 2

AnswEr :

We're given with certain values, that is,

⠀⠀⠀⠀⠀⠀⠀⠀⠀

• Velocity of cyclist, Vc = 20 m/s

• Velocity of Bus, Vb = 0 m/s [As the Bus has started from rest ]

• Acceleration of Cyclist, Ac = 0 m/s²

• Acceleration of Bus, Ab = 2 m/s²

• Distance between Cyclist & Bus, S = 96 m[Seperation]

⠀⠀⠀⠀⠀⠀⠀⠀⠀

Reference of Diagram :

⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\setlength{\unitlength}{2mm}\begin{picture}(0,0)\thicklines\put(0,0){\line(3,0){7cm}}\put(3,5){\circle{2}}\put(2,0){\line(1,2){3mm}}\put(4,0){\line(-1,2){3mm}}\put(2.95,2){\line(0,3){5mm}}\put(1.69,3){\line(3,0){5mm}}\put(4,2){ \sf\footnotesize \xrightarrow{Speed \;of\;cyclist\:=20 \; ^m\!/_s} }\put(25,2){\line(0,3){0.7cm}}\put(25,2){\line(3,0){1cm}}\put(30,2){\line(0,3){0.7cm}}\put(25,5.5){\line(3,0){1cm}}\put(26,1){\circle{2}}\put(29,1){\circle{2}}\put(5,-1){ {\underbrace{\qquad\qquad\qquad\qquad\qquad\quad} }}\put(13,-4){\sf\footnotesize 96m}\put(31,2){\sf\footnotesize Bus_{\sf acc}=2\;\sf^m\!/s^2 }\end{picture}$

In order to solve this kind of questions, we need to know the relations between the following 2 things.

1. Relative Velocity

2. Relative Acceleration

With respect to their relative separation !

After getting these values, we'll use the kinematic equation to get the actual time taken by the cyclist to overtake the bus.

↬ Relative Velocity = Velocity(cyclist) - Velocity (Bus)

↬ Vcb = Vc - Vb

↬ Vcb = 20 - 0

↬ Vcb = 20 m/s

Now,

↬ Relative Acceleration = Acceleration(cyclist) - Acceleration(Bus)

↬ Acb = Ac - Ab

↬ Acb = 0 - 2

↬ Acb = -2 m/s²

Also, Relative Separation is given which is 96 m.

By using the kinematic equation :

$\longrightarrow \tt S = ut + \dfrac{1}{2}at^2$

$\longrightarrow\tt 96 = 20t - \dfrac{1}{2}2t^2$

$\longrightarrow\tt t^2 - 20t + 96 = 0$

$\longrightarrow\tt t^2 - 8t - 12t + 96 = 0$

$\longrightarrow\tt t(t - 8) -12(t - 8) = 0$

$\longrightarrow\tt (t-8)(t-12) = 0$

$\longrightarrow\tt {\boxed{\tt{t = 8}}} \: or \: { \boxed{\tt{t = 12}}}$

$\therefore$ After 8s the cyclist will overtake the bus.

Also, the same bus will overtake the cyclist within 12s.