Two cars 1 and 2 move with velocities v1 and v2 respectively on a straight road in the same direction. When the cars are separated by a distance d, the driver of the car 1 applies brakes and the car moves with uniform retardation a1. Simultaneously, the car 2 starts accelerating with a2. If v1>v2, find the minimum initial separation between the cars to avoid collision between them.
Two cars 1 and 2 move with velocities v1 and v2 respectively on a straight road in the same direction. When the cars are separated by a distance d, the driver of the car 1 applies brakes and the car moves with uniform retardation a1. Simultaneously, the car 2 starts accelerating with a2. If v1>v2, find the minimum initial separation between the cars to avoid collision between them.
The correct option is D
Method 1
Let us assume that the cars undergo displacement S1 and s2 after a time t from the instant or braking.
From figure, s1−s2=d.....(i),
Where s1=v1t−12a1t2....(ii)
Using equations(i), (ii) and (iii), we have (v1t−12a1t2)−(v2t−12a1t2)=d
Or, a1+a2)t2−2(v1−v2)t+2d=0
This is quadratic equation in t, solving for t
That gives t=(v1+v2)2−2(a1+a2)da1+a2
If cars do not collide, b2−4ac<0⇒(v1−v2)2−2(a1+a2)d<0
(v1−v2)22(a1+a2)<d Hence dmin=(v1−v2)22(a1+a2)
Method 2.
Relative velocity approach:
Using v2rel=u2rel+2arel.sre we can obtain the same result.
Substitute: vrel=v′12=v′1−v′2, urel=v1−v2,
arel=a12=(−a1)−a2=−(a1+a2) and srel=S12=−d, to obtain (v′1−v′2)2=(v1−v2)2−2(a1+a2)d
Since v′1<v′2 to avoid collision, we have d=(v1−v2)2−(v1−v2)22(a1+a2)
Maximum distance d for collision is possible is possible when v′1=v′2 then, dmax=(v1−v2)22(a1+a2)
Method 1
Let us assume that the cars undergo displacement S1 and s2 after a time t from the instant or braking.
From figure, s1−s2=d.....(i),
Where s1=v1t−12a1t2....(ii)
Using equations(i), (ii) and (iii), we have (v1t−12a1t2)−(v2t−12a1t2)=d
Or, a1+a2)t2−2(v1−v2)t+2d=0
This is quadratic equation in t, solving for t
That gives t=(v1+v2)2−2(a1+a2)da1+a2
If cars do not collide, b2−4ac<0⇒(v1−v2)2−2(a1+a2)d<0
(v1−v2)22(a1+a2)<d Hence dmin=(v1−v2)22(a1+a2)
Method 2.
Relative velocity approach:
Using v2rel=u2rel+2arel.sre we can obtain the same result.
Substitute: vrel=v′12=v′1−v′2, urel=v1−v2,
arel=a12=(−a1)−a2=−(a1+a2) and srel=S12=−d, to obtain (v′1−v′2)2=(v1−v2)2−2(a1+a2)d
Since v′1<v′2 to avoid collision, we have d=(v1−v2)2−(v1−v2)22(a1+a2)
Maximum distance d for collision is possible is possible when v′1=v′2 then, dmax=(v1−v2)22(a1+a2)
