Question

An object initially at rest experiences an acceleration of 2.20 m/s2 on a level surface. Under these conditions, it travels 5.00 m . Let’s designate the first 2.50 m as phase 1 with a subscript of 1 for those quantities, and the second 2.50 m as phase 2 with a subscript of 2. The times for traveling each phase should be related by which condition?

Answer 1

Answer:

t2  = Time for whole trip - time t1  for first phase trip = 0.6 sec

Explanation:

The object starts from rest;

Vi=0 m/s

a = 2.20 m/sec2

S= distance traveled

t= time

Here we can first  calculate time expression in terms of distance S;

We know

$S= Vit + \frac{1}{2}at^{2}$

Putting values

==>S= 0+$\frac{1}{2}$ ×2.2×$t^{2}$

==> $t^{2}$ = 2S/2.2

==> t = 0.953462589$\sqrt{S}$

Lets suppose it completes the trip without 'phases' then S here is 5 meters as a whole ;

==> t = 0.953462589 × $\sqrt{5}$

==> t = 0.953462589 × 2.236067977= 2.132007163 Sec

Now let's consider phases

Phase 1:

Let S = S1 = 2.5 m

t=t1

==> t1 = 0.953462589 × $\sqrt{2.5}$ =  1.507556723 sec

So for the second phase the time t2  = Time for whole trip - time t1 for first phase trip

==> t2 = 2.132007163 - 1.507556723 = 0.62445044 = 0.6 sec