Find the acceleration and time?

Question

A speedboat increases its speed uniformly from 20 m/s to 25 m/s in a distance of 200 m.

(a) Find the magnitude of its acceleration.

(b) Find the time it takes the boat to travel the 200 m distance.

Answer 1

v= v0 + at
25 = 20 + at
t = 5/a

d= 1/2a t^2 + v0t
200 = 1/2a t^2 + 20t
200 = 1/2 a (5/a)^2 + 20(5/a)
200 = 25/2a + 100/a = 225/2a
a = 225/400 = 0.5625 m/s^2

t = 5/0.625 = 8.889 s

Answer 2

a) The magnitude of the acceleration = 0.5625 m/s^2.

b) The time taken = 8.888...secs = 8 8/9 secs.

["catartur's" initial answer was unfortunately incorrect. In between the following two lines:

200 = 1/2 a (5/a)^2 + 20(5/a)
200 = 25/a + 100/a = 125/a,

he dropped the "1/2" in the first term on the RHS. He has since corrected it.]

I used the following equations from the simple kinematics of uniform acceleration:

1. vf^2 = v0^2 + 2 a (sf - so), where "subscripts" f, 0 mean "final," "initial." *** [See footnote.]

2. vf = v0 + a t.

Applying these two equations sequentially:

1. 625 = 400 + 2 (200) a, thus a = 225 / 400 = 0.5625 m/s^2.

2. 25 = 20 + 0.5625 t, thus t = 5 / 0.5625 = 8.88... secs or 8 8/9 secs.

Live long and prosper.

*** [FOOTNOTE : Note that this is the MOST DIRECT way of linking the givens, i.e. speed -- distance -- acceleration. One can also recognize in it the conservation of energy (per unit mass) in a uniform gravitational field with ' a ' = ' g .' It is always worthwhile to have this additional kinematical arrow in one's mathematical quiver!] ***