Just as a car starts to accelerate from rest with an acceleration of 2ms^-2,a bus moving with constant speed of 10ms^-1 passes it in a parallel lane a)how fast will the car be going when it overtakes the bus
@ganeshie8
so the bus is going at 10m/s speed, can you write the equation for its distance ?
\[\large s = ut + \dfrac{1}{2}at^2\] seen this before ?
yes
\(u = 10\) \(a = 0\) plugin
you get : \[\large s = 10t + \dfrac{1}{2}(0)t^2\] \[\large s = 10t ~~~~~~~~~~\color{Red}{(1)}\]
s=10(t)+0
thats the equation for distancce travelled by the bus as time goes on
we need to find at what time the car covers this distance and overtakes the bus
lets write a distance equation for `car` also : \(u = 0\) \(a = 2\) plugg them in distance equation
you get : \[\large s = (0)t + \dfrac{1}{2}(2)t^2\] \[\large s = t^2~~~~~~~~~\color{Red}{(2)} \]
want to clarify this ..., a = 0 coz constant velocity
exactly !
velocity is not changing, so there is no acceleration
solve \(\large \color{red}{(1)}\) and \(\large \color{red}{(2)}\) for \(t\)
t=10
yes ! so the car overtakes the bus exactly after 10 seconds
that means the car accelerates for 10 seconds before overtaking the bus, right ?
t = 10 u = 0 plug these in : \(\large v = u+at\) to find the speed at which the car overtakes the bus
v=0+a(10)
Oh yes, and you're given a = 2, plug that also
v=20ms^-1
Yep ! so the car overtakes the bus exactly after 10 seconds and the speed of car when overtaking is 20 ms^-1
Thanks @ganeshie8
np :) did u get why we have equated the equations \(\large \color{red}{(1)}\) and \(\large \color{red}{(2)}\) earlier ?
to find time..
we equated them because, the `distances travelled by both car and bus are equal` when the car overtakes the bus
Okay, understood
good :)
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