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Physics 17 Online
OpenStudy (eric_d):

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

OpenStudy (eric_d):

@ganeshie8

ganeshie8 (ganeshie8):

so the bus is going at 10m/s speed, can you write the equation for its distance ?

ganeshie8 (ganeshie8):

\[\large s = ut + \dfrac{1}{2}at^2\] seen this before ?

OpenStudy (eric_d):

yes

ganeshie8 (ganeshie8):

\(u = 10\) \(a = 0\) plugin

ganeshie8 (ganeshie8):

you get : \[\large s = 10t + \dfrac{1}{2}(0)t^2\] \[\large s = 10t ~~~~~~~~~~\color{Red}{(1)}\]

OpenStudy (eric_d):

s=10(t)+0

ganeshie8 (ganeshie8):

thats the equation for distancce travelled by the bus as time goes on

ganeshie8 (ganeshie8):

we need to find at what time the car covers this distance and overtakes the bus

ganeshie8 (ganeshie8):

lets write a distance equation for `car` also : \(u = 0\) \(a = 2\) plugg them in distance equation

ganeshie8 (ganeshie8):

you get : \[\large s = (0)t + \dfrac{1}{2}(2)t^2\] \[\large s = t^2~~~~~~~~~\color{Red}{(2)} \]

OpenStudy (eric_d):

want to clarify this ..., a = 0 coz constant velocity

ganeshie8 (ganeshie8):

exactly !

ganeshie8 (ganeshie8):

velocity is not changing, so there is no acceleration

ganeshie8 (ganeshie8):

solve \(\large \color{red}{(1)}\) and \(\large \color{red}{(2)}\) for \(t\)

OpenStudy (eric_d):

t=10

ganeshie8 (ganeshie8):

yes ! so the car overtakes the bus exactly after 10 seconds

ganeshie8 (ganeshie8):

that means the car accelerates for 10 seconds before overtaking the bus, right ?

ganeshie8 (ganeshie8):

t = 10 u = 0 plug these in : \(\large v = u+at\) to find the speed at which the car overtakes the bus

OpenStudy (eric_d):

v=0+a(10)

ganeshie8 (ganeshie8):

Oh yes, and you're given a = 2, plug that also

OpenStudy (eric_d):

v=20ms^-1

ganeshie8 (ganeshie8):

Yep ! so the car overtakes the bus exactly after 10 seconds and the speed of car when overtaking is 20 ms^-1

OpenStudy (eric_d):

Thanks @ganeshie8

ganeshie8 (ganeshie8):

np :) did u get why we have equated the equations \(\large \color{red}{(1)}\) and \(\large \color{red}{(2)}\) earlier ?

OpenStudy (eric_d):

to find time..

ganeshie8 (ganeshie8):

we equated them because, the `distances travelled by both car and bus are equal` when the car overtakes the bus

OpenStudy (eric_d):

Okay, understood

ganeshie8 (ganeshie8):

good :)

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