Answer:
Explanation:.
Given that
First car velocity is 2m/s
Second car velocity is 9m/s.
At a certain time the first car is 8m from intersection
And at the same time second car is 6m from intersection.
The rate of change of distance, i.e dx/dt, which is the speed of the car.
Using Pythagoras theorem, their distance apart is given as
Z²=X²+Y²
Z²=6²+8²
Z²=36+64
Z²=100
Z=√100
Z=10m
Let assume the direction,
Let assume the first car is moving in positive x direction, then
dx/dt=2m/s
And also second car will be moving in negative y direction
dy/dt=-9m/s
Now, to know dz/dt, let use the Pythagoras formulae above
x²+y²=z²
differentiate with respect to t
2dx/dt+2dy/dt=2dz/dt
Divide through by 2
dz/dt=dx/dt+dy/dt
dz/dt=2-9
dz/dt=-7m/s
The rate of change of distance between the two body is -7m/s
Option B is correct
Answer:
Rate of change of the distance between the cars = -7 m/s
Explanation:
Let the distance between the first car and the intersection be p = 8 meters
Let the distance between the second car and the intersection be q = 6 meters
velocity of the first car, dp/dt = -2 m/s
velocity of the second car, dq/dt = -9 m/s
We can get the distance between the two cars using pythagora's theorem
s² = p² + q²...................................(1)
s² = 8² + 6²
s = √(8² + 6²)
s = 10 m
Differentiating equation (1) through with respect to t
2s ds/dt = 2p dp/dt + 2q dq/dt
(2*10*ds/dt) = (2*8*(-2)) + (2*6*(-9))
20 ds/dt = -32 - 108
20 ds/dt = -140
ds/dt = -140/20
ds/dt = -7 m/s
