Question

A police cruiser, approaching a right-angled intersection from the north, is chasing a speeding car that has turned the corner and is now moving straight east. When the cruiser is $0.6$ km north of the intersection and the car is $0.8$ km to the east, the police determine with radar that the distance of the car is increasing at $20$ km/h. Suppose that the cruiser is moving at $60$ km/h at the instant of measurement.

$\frac{d^{2}s}{d{t}^2}$ equal to at the moment in question if $\frac{d^{2}x}{d{t}^2}=-40$ and $\frac{d^{2}y}{d{t}^2}=50$(in km/h${}^2$)

• A. $6400$
• B. $6450$
• C. $6200$
• D. $6000$

Answer 1

Answer: (B)

$6450$

At the instant in question $x=0.8,y=0.6,$
$\frac{dy}{dt}=-60km/h$, $\frac{ds}{dt}=20km/h$, $s^2=x^2+y^2$.
Substitnting the given values, we get
$20=0.8\frac{dx}{dt}-36\Rightarrow \frac{dx}{dt}=70$
$\frac{d^{2}s}{d{t}^2}=\frac{1}{\sqrt{x^2+y^2}}[(1-\frac{x}{x^2+y^2}){\left(\frac{dx}{dt}\right)}^2+(1-\frac{y}{x^2+y^2}){\left(\frac{dy}{dt}\right)}^2+x\frac{d^{2}x}{d{t}^2}+y\frac{d^{2}y}{d{t}^2}-\frac{2xy}{x^2+y^2}\frac{dx}{dt}\frac{dy}{dt}]$
Substitute the given value $\frac{d^{2}s}{d{t}^2}=6450$.