# [Physics] Intercept a moving object

homework-and-exerciseskinematics

Object A can move at 50km/h, wants to intercept object B (currently $15^{\circ}$, east of north from A) moving at 26km/h, $40^{\circ}$ east of north. What angle should A take to intercept B? AB is 20km apart

Choose x axis along 20km distance.

$26t \sin{(40-15)} = 50t \sin{\theta}$

$\theta = \sin^{-1}{\frac{11}{50}} = 12.7$

$15 + 12.7 = 27.7$

I took a different approach and used $\cos$ and got a different answer … why is that?

$26t \cos{(40-15)} = 50t \cos{\theta}$

Updated

## Using x-axis along AB

Taking the x-axis along AB yields

$$50 t \sin (\theta-15^\circ) = 26 t \sin(40^\circ)$$ $$\sin (\theta-15^\circ) = 0.52 \sin(40^\circ)$$

$$\theta = 34.527^\circ$$

$$\cos (\theta-15^\circ) = \sqrt{1-\sin^2 (\theta-15^\circ) }$$

and the y-axis perpendicular to AB

$$50 t \cos(\theta-15^\circ) = 20 + 26 t \cos(40^\circ)$$

$$t = 0.735$$

## Using x-axis along AC (interception pt)

taking y-axis perpendicular to AC

$$50 t = 20 \cos(\theta-15^\circ)+26 t \cos(55^\circ-\theta)$$ $$t = \frac{20 \cos(\theta-15^\circ)}{50 - 26 \cos(55^\circ-\theta)}$$

taking x-axis along AC $$20 \sin(\theta-15^\circ)=26 t \sin(55^\circ-\theta)$$

which when expanded you need to solve an equation of the form $$A\cos \theta + B \sin \theta = C$$ for $\theta$ with the same results as above.