A has velocity 4 m/s towards B. Velocity of B is 1 m/s towards A. The separation between them is 20 m. Displacement of A when they meet
Answers
Answer:
Distance traveled by car A=S A
Distance traveled by car B=S B
According to formula S=ut+ 21
at
2SA =0t+
2
1 ×2t2SB
=10t+ 21 ×0t2
Car A will overtake car B when
S A =SB+100
Putting the value S Aand S B
+2=10t+100
+2−10t−100=0
After solving for t, we get
t=16.18sec possible
t=−6.10sec not possible
Answer:
Using the concept of relative velocity
Let us assume the velocity of A w.r.t. B. To do this, we plot the resultant velocity,
V
→
AB
V
→
AB
=
V
→
A
−
V
→
B
=
V
→
A
+(−
V
→
B
)
As the accelerations of both the cars is zero, so the relative acceleration between them is also zero. Hence, the relative velocity will remain constant. So the path of A with respect to B will be straight line and along the direction of relative velocity of A with respect to B. The shortest distance between A and B is a perpendicular from B on the line of motion of A with respect to B.
From the figure
tanθ=
V
A
V
B
=
20
15
=
4
3
....(i)
This θ is the angle made by the resultant velocity vector ∣
V
→
AB
∣ with the x-axis.
Also we know that from Fig.
OC=
500
x
=
4
3
From equation (i) and (ii), we get x = 375 m.
∴AB=OB−OC=400−375=25m
But the shortest distance is BP.
From diagram, it is clear that BP=BCcosθ=25×
5
4
∴BP=20m
