Question

In a car race 'car A' takes 't' less than 'car B' and passes the finishing point with velocity 'v' more than the velocity with which 'car B' passes the point, assuming that the cars start from rest and travels with constant acceleration 'a1 & a2', show that v = t root a1a2.

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Answer 1

Answer:

v = t √a1a2

Step-by-step explanation:

Let say Car A takes  t1  time  

Then Velocity of Car A  =  0 + a1t1

Car B takes  t1+t  time  

Then Velocity of Car B  =  0 + a2(t1 + t)

a1t1 = a2(t1 + t) + v

=> t1(a1 - a2) = a2t + v

=> t1 = (a2t + v)/(a1 - a2)

Distance covered by Car A = (1/2)a1t1²

Distance covered by Car B = (1/2)a2(t1 + t)²

(1/2)a1t1² = (1/2)a2(t1 + t)²

=> a1t1² = a2(t1 + t)²

=> a1( (a2t + v)/(a1 - a2))²  = a2((a2t + v)/(a1 - a2)  + t)²

=> a1(a2t + v)² = a2(a2t + v + a1t - a2t)²

=>  a1(a2t + v)² = a2(v + a1t)²

=> a1(a2²t² + v² + 2a2tv) = a2(v² + a1²t² + 2va1t)

=> a1a2²t² + a1v² + 2a1a2tv = a2v² + a2a1²t² + 2a1a2tv

=>  a1a2²t² + a1v² = a2v² + a2a1²t²

=> v²(a1 - a2)  = a1a2t²(a1 - a2)

=> v² = a1a2t²

=> v = t √a1a2

Answer 2

Answer:

Step-by-step explanation:

Consider that A takes t1 second, then according to the given problem, B will take (t1+t) seconds. Further let v1be the velocity of B at finishing point, then velocity of A will be (v 1 +v). Writing equations of motion for A and B

V1+v=a1 t ....(i)

v1=a 2

(t2+t) ....(ii)

From equations (i) and (ii), we get

v=(a1−a2 )t1−a2 t .....(iii)

Total distance travelled by both the cars is equal

S A=S B

⇒ 21 a1 t12

= 21a2 (t1+t)2

⇒t 1 = a1 − a2a2 t

Substituting this value of t1

in equation (iii), we get

v=( a1 a2 )t