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

The two cars with given accelerations and an initial distance of 1400 meters between them when starting from rest will meet after 20 seconds.

Step-by-step explanation:

To solve for the time at which the two cars meet, we must consider the accelerations of both cars and the initial distance between them. Assuming car A is moving with a positive acceleration of 11 m/s2 and car B is moving with a negative acceleration of -4 m/s², we need to find a common point in time where they both cover the total distance of 1400 m when starting from rest.

Let the time taken for the cars to meet be denoted by 't'. For car A, the displacement (sA) is given by the formula sA = 0.5 * aA * t², and for car B, the displacement (sB) is given by sB = 0.5 * aB * t². As they are moving towards each other, the sum of their displacements sA+sB should equal the initial separation distance which is 1400m.

This gives us the equation 0.5 * 11 * t2 + 0.5 * (-4) * t² = 1400. Simplifying, we get 3.5 * t² = 1400, and solving for 't' gives us t2 = 400, so t = 20 seconds. Hence, both cars will meet after 20 seconds.

Answer 2

13.7 s

Step-by-step explanation:

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

`x_A (t) = (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+(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,

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