Question

Two guns are placed 900 ft away from each other. If the gun on the left fires its bullet at 1200 ft/s and the gun on the right fires at 1500 ft/s, how far from the middle will the bullets meet?

• A. 50 ft to the left of the middle
• B. 50 ft to the right of the middle
• C. 100 ft to the left of the middle
• D. 100 ft to the right of the middle

Answer 1

The correct option is A 50 ft to the left of the middle

${\text{Let's say that the bullets collide}}^{\prime }{t}^{\prime }\text{seconds after being fired.}$

$Gunontheleft:$
$\text{Bullet speed}=1200ft/s$
${\text{Distance traveled by the bullet in}}^{\prime }{t}^{\prime }\text{seconds}=1200\times t=1200tft$

$Gunontheright:$
$\text{Bullet speed}=1500ft/s$
${\text{Distance traveled by the bullet in}}^{\prime }{t}^{\prime }\text{seconds}=1500\times t=1500tft$

$\text{Total distance}=900ft$

$⟹1200t+1500t=900$
$⟹2700t=900$
$⟹t=\frac{900}{2700}=\frac{1}{3}$

$\text{So, the bullets meet in}\frac{1}{3}\text{seconds.}$

$\text{Distance covered by the bullet from the}\text{left in}\frac{1}{3}\text{seconds}=1200ft/s\times \frac{1}{3}s=\frac{1200}{3}ft=400ft$

$\text{Comparing this to the point in the middle,}\text{we find:}$


$\text{Hence, the bullets meet 50 ft to the left}\text{of the middle.}$