Question

Two cars are heading towards one another . Car a is moving with an acceleration of 11. And carb is moving at -4 m/s^2. The cars are at rest and seperated witha distance of s=1400m. What time do the cars meet?

Answer 1

Answer:

13.7 s

Explanation:

The position at time t of car A can be written as follows:

$x_A (t) = \frac{1}{2}a_At^2$

where

$a_A = 11 m/s^2$ is the acceleration of car A

The position of car B instead can be written as

$x_B(t) = d+\frac{1}{2}a_B t^2$

where

$a_B = -4 m/s^2$ is the acceleration of car B

d = 1400 m is the initial separation between the cars

The two cars meet when

$x_A = x_B$

Using the two equations above,

$\frac{1}{2}a_A t^2 = d + \frac{1}{2}a_B t^2\\\frac{1}{2}t^2 (a_A - a_B) = d\\t=\sqrt{\frac{2d}{a_A-a_B}}=\sqrt{\frac{2(1400)}{11-(-4)}}=13.7 s$