What was the total displacement x for the complete trip Answ

What was the total displacement x for the complete trip? Answer in units of m.

A record of travel along a straight path is as follows: (a)Start from rest with constant acceleration of 2.28m/s^2 for 19.2;(b) constant velocity for the next 0.894 min; (c) constant negative acceleration of -10.3 m/s^2 for 3.75s.

any suggestion will be appreciated!!. thanks .

Solution

The entire trip has 3 parts , (a) , (b) and (c)

a) The constant acceleration of the record = 2.28 m/s^2

for 19.2 seconds . Time units assumed in secods.Therefore the displacement ,s = u+(1/2) at^2 , where u is the initial velocity and a is the constant acceleration and t is time in seconds.Given u=0, a =2.28m/s^2 and t=19.2secs . So, s= 0*(19.2)+(1/2)(2.28)(19.2)^2=420.2496 m

b)The velocity of the record at the end of 19.2 secs =u+at =0+2.28*19.2 =43.776m .So at this constant velocity for 0.894 *60 secs the record displaces further by =43.776*0.894*60=2348.1446m

c)Now u=v=43.776, a= -10.3m/s^2, t=3.75s Therefore the displacement is given by the formila ut +(1/2)at^2 = 43.776*3.75+(1/2)(-10.3)3.75^2= 91.7381m

Therefore the total displacement in (a) , (b) and (c) =420.2496+2348.1446+91.7381=2860.1323 meters aproximately.

This will help u.