Question

Two cars are driving towards an intersection from perpendicular directions. The first car's velocity is 2 meters per second and the second car's velocity is 9 meters per second. At a certain instant, the first car is 8 meters from the intersection and the second car is 6 meters from the intersection. What is the rate of change of the distance between the cars at that instant (in meters per second)? Choose 1 answer:
a -10
b -7
© -8.4

Answer 1

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

Answer 2

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