plzz solve this...

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??

5 Answers

62
Lokesh Verma ·

seems 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

62
Lokesh Verma ·

The 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..

62
Lokesh Verma ·

Thus, 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?

1
pinky ·

i cud solve the first part..tht's quite simple...but can u plzz explain me how to solve the second part of the prob.??

62
Lokesh Verma ·

Second 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!

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