Answer:
30 m/s
Step-by-step explanation:
Let's say the distance from the first car to the intersection is x, and the distance from the second car to the intersection is y.
The distance between the cars can be found with Pythagorean theorem:
d² = x² + y²
Taking derivative with respect to time:
2d dd/dt = 2x dx/dt + 2y dy/dt
d dd/dt = x dx/dt + y dy/dt
We know that x = 200, dx/dt = -25, y = 150, and dy/dt = -50/3.
To find dd/dt, we still need to find d.
d² = x² + y²
d² = (200)² + (150)²
d = 250
Plugging everything in:
250 dd/dt = (200) (-25) + (150) (-50/3)
dd/dt = -30
The cars are approaching each other at a rate of 30 m/s at that instant.
The cars are approaching each other at 30 m/sec
The rate of approach is the value at which the distance between the two cars is reduced with respect to time.
The Pythagoras theorem says that in a right-angled triangle the sum of squares of the base(B) and the perpendicular(P) is equal to the square of the hypotenuse(H).
H² = B² + P²
In the given problem the two cars along with the intersection point make a right-angled triangle, with a right angle at the point of intersection.
If the distance between the two cars is taken to be D, the distance between the first car and the point of intersection as X, and the distance between the second car and the point of intersection as Y, then by Pythagoras Theorem we get
D² = X² + Y² ...........(i)
To find the rate, we differentiate this equation with respect to time T,
2D dD/dT = 2X dX/dT + 2Y dY/dT
⇒ D dD/dT = X dX/dT + Y dY/dT ............(ii)
By the problem, we have
X = 200m, Y = 150m, dX/dT = -25m/sec, dY/dT = - 50/3 m/sec.
Substituting values of X and Y in (i), we get
D² = 200² + 150² = 40000 + 22500 = 62500
⇒ D = √62500 = 250m
Now substituting all values in (ii), we get,
250 dD/dT = 200 * (-25) + 150 * (-50/3) = -5000 - 2500 = -7500
⇒ dD/dT = -7500/250 = -30 m/sec
∴ The cars are approaching each other at 30 m/sec.
Learn more about the Pythagoras Theorem and Rate at
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