A body moves with constant acceleration a = 14 cm/s^2 . At…

A body moves with constant acceleration $a = 14\ \mathrm{cm/s^2}$. At time $t_1 = 3\ \mathrm{s}$ after the start of the motion, its speed is $v_1 = 82\ \mathrm{cm/s}$. What was its initial speed, and what distance does it cover in the first $t_2 = 10\ \mathrm{s}$?

Explanation of Method

Initial speed and distance from one measurement.

Step 1: Initial speed

With constant acceleration the speed grows by the same amount every second: $v_1 = v_0 + a t_1 \;\Longrightarrow\; v_0 = v_1 - a t_1$.

\[ v_0 = 82 - 14 \cdot 3 = 82 - 42. \]

\[v_0 = 40\ \mathrm{cm/s}\]

Step 2: Distance in the first 10 s

\[ s = v_0 t_2 + \tfrac12 a t_2^2 = 40 \cdot 10 + \tfrac12 \cdot 14 \cdot 10^2 = 400 + 700. \]

\[s = 1100\ \mathrm{cm} = 11\ \mathrm{m}\]

Check with the average speed

Speed at 10 s: $ v = 40 + 14 \cdot 10 = 180\ \mathrm{cm/s} $

Average speed: $ \tfrac{40 + 180}{2} = 110\ \mathrm{cm/s} $, so $ s = 110 \cdot 10 = 1100\ \mathrm{cm}\ \checkmark $

$v\ (\mathrm{cm/s})$ $t\ (\mathrm{s})$ $0$ $3$ $5$ $10$ $40$ $82$ $180$ $v_0 t = 400\ \mathrm{cm}$ $\tfrac{1}{2}at^2 = 700\ \mathrm{cm}$ $(3\ \mathrm{s},\ 82\ \mathrm{cm/s})$ $180$

from the initial speed
from the acceleration

The area under the speed–time line from 0 to 10 s is the distance: a rectangle (400 cm) plus a triangle (700 cm) = 1100 cm.

Result

\[\boxed{\,v_0 = 40\ \mathrm{cm/s}\,}\]

\[\boxed{\,s = 1100\ \mathrm{cm} = 11\ \mathrm{m}\,}\]