1i cud solve the first part..tht's quite simple...but can u plzz explain me how to solve the second part of the prob.??
two motor cars start from a point A simultaneously and reach at point b after 2hrs. The first car travelled half of the dist. ata speed of 30km/hr & the other half at a speed of 60 km/hr. The second car covers the entire dist. with a const. acceleration. At wat instant of time the speeds of both the cars are same? Will one of them overtake the other??
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5 Answers
62seems sort of simple..
So we will take the time for which it travels with 30km/hr as T hrs
So the time for which it travels with 60km/hr as 2-T
Distance for both is same
30T=60(2-T)
T=2(2-T)
3T=4
T=4/3
So the velocity is given by
v=30 t<=4/3
v=60 for t>4/3
62The distance for the first car
= 4/3 . 30 *2= 80 Km.. = distance by the 2nd car..
So, 80=1/2 a t2 (where t=2)
So a=40km/hr2
Now velocity of this car at time T will be
aT = 40T
40T=30 when T=3/4 hrs
and will be 60 when 40T=60 T=3/2 Hrs..
62Thus, there will be 2 such times at which such a thing will happen!
The second part is all about finding displacement as a function of time..
keep doing it in the way i did this one.. it should be simple enuf?
162Second part
The displacemet will be integral of veloicty
First Car: S=Vt
For t<4/3, S=30t
For t>4/3, S=40+(t-4/3).60
S=60t-40
For the second Car,
S=1/2 at2
Try case t<4/3
1/2 at2=30t
2t2=3t
t=0 or t=3/2 whcih does not lie in this region.. so no valid solution for this case except the initial point.
For case t>4/3
1/2 at2=60t-40
20t2=60t-40
t2=3t-2
t2-3t+2 = 0
t=1 or t=2
Only t=2 lies in this region.
So the two cars will meet again at t=2.
So the cars will never overtake each other! (For that S has to be equal at some point in time!)
Here these two points happened to be the beginning and the end points!