Two cars are heading towards one another car a is moving...

Two cars are heading towards one another. Car A is moving with an acceleration of aA = 4 m/s2. Car B is moving with an acceleration of aB = -14 m/s2. The cars are initially at rest and separated by a distance d = 4400 m on the x-axis. (a) At what time, in seconds, do the cars meet? t = (b) What is the displacement, in meters, of Car A? xA = (c) What is the displacement, in meters, of Car B? xB = (d) Now assume that both cars start with an initial velocity. Car A's initial velocity is v0A = 47 m/s and Car B's initial velocity is vB = -26 m/s. At what time, t2 in seconds, do they meet? t2 =

Transcript

00:01 Hello everyone in this question it is given that acceleration of card a a suffix capital a equals to 4 meter per second square and acceleration of card b a suffix b equals to minus 14 meter per second square and distance between the card b equals to given 4400 meter so here we need to find out at what time two cards meet each other and this is the first part of the question means part a so here initial velocity both cards ua equals to u b equals to 0 so here we are using an emitting equation so using the equation of motion which is x of a equal to ua t plus half a t squared n x b equals to u b t plus half a suffix b t squared so now here putting the values we will got the value of x a equals to half 4 t squared and the value of xb equals to half minus 14 t squared so for the cards to meet at xa equals to x b plus 4400 as the cards are initially separated by 4 ,400 meter.

01:43 So therefore from here we can retain as the value of x is half 4 t squared is equal to half minus 14 t squared plus 4400.

01:58 So further solving this here, we will got 18 t squared equals to 8800.

02:07 So from here we will got t equals to 48800.

02:08 So from here we will got t equals to 4808 ,000.

02:13 Point 8 .89 means here is square root of 488 .89 so further solving this we will got the value of t equals to 22 .11 second so time at which both cars meet is t equals to 32 .11 second so now coming to the part b so for part b we have to find out the displacement of card a so final position of car a we can return as using kinetic equation xa equals to u of a multiplied by t plus half a suffix a t squared so here it is given that u a equals to 0 a suffix a we already calculated which value is 4 meter per second square and the time t equals to 22 .11 second so now putting the values therefore we will got x of a equals to 0 plus half multiplied by 4 multiplied by 22 .11 whole square.

03:35 So, for solving this, we will got the value of xa equals to 977 .78 meter...

Question

Please give Ace some feedback