Question

An object accelerates from rest to velocity of 38m/s over a distance of 12m what was its acceleration ?

Answer 1

answer

60.1667 $\frac{m}{s^2}$

explanation

there are four main kinematic formulas that we use in physics

$v_{f}= v_{o}+at\\x =\frac{t(v_{f}+v_o)}{2} \\x = v_ot + \frac{at^2}{2} \\{v_{f}}^2 = {v_o}^2 + 2ax$

$v_f$ = final velocity

$v_o$ = initial velocity

a = acceleration

t = time

x = displacement

we use an equation based on what information we have

in this question, we have initial velocity (0), final velocity, (38), displacement (12), and we want to find acceleration

so we use the equation that includes $v_o, v_f, x, a$

which is this equation ${v_{f}}^2 = {v_o}^2 + 2ax$

rearrange to solve for a

${v_{f}}^2 = {v_o}^2 + 2ax\\{v_{f}}^2}-{v_o}^2 = 2ax\\a = \frac{{v_f}^2-{v_o}^2}{2x}$

plug in values

$a = \frac{{v_f}^2-{v_o}^2}{2x}\\\\a = \frac{{38}^2-{0}^2}{2*12}\\\\a = \frac{1444}{24} \\\\a = 60.1667$