Question

A car slows from a velocity of 26m/s to 18m/s in a distance of 52 meters. What was the cars acceleration

Answer 1

-3.4 by use v^2=u^2+2as

Answer 2

Recall that

v² - u² = 2 a ∆x

where u and v denote initial and final velocities, respectively; a is acceleration; and ∆x is the distance traveled.

Then we get

(18 m/s)² - (26 m/s)² = 2 a (52 m)

a = ((18 m/s)² - (26 m/s)²) / (104 m)

a ≈ -3.4 m/s²