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 0.894min; (c)constant negative acceleration of -10.3 m/s^2 for 3.75 s.

Solution

Displacement for the object which starts with 0 vel and constant acceleration a for 19.2 secs.Here the final velocity at the end 19.2 sec =initial velocity+acceleration* time=0+2.28*19.2 =2.28*19.2.

So the average speed ( initial velocity+final velocty)/2=(0+uniform acc*time)/2 = (2.28*19.2)/2.

So, the displacement in19.2 secs= avg speed*time =[(2.28*19.2)/2]*19.2=420.2496m

The displance with the constant velocity of 2.28*19.2 for 0.894 min(=0.894*60 secs)= 2.28**894*60=2348.14464

The particle is now at the initial velocity 2.28*19.2m/sec and retards at 10.3/s^2 for 3.75 seconds. So, it displacement s could be calculated by :s= ut+(1/2)at^2, where u= initial velocity = 2.28*19.2m/s, and a=acceleration=-10.3m/s^2 and t=time of movement= 3.75s. Therefore, s=(2.28*19.2)(3.75)+(1/2)(-10.3)(3.75^2)=91.7381

Since the entire displacements are ina straight line, the total displacement =   420.2496+2348.1446+91.7381=2860.1323m.