Question

A bus traveling at 30 m/s accelerates at a constant rate of 2.4 m/s2 over a distance of 60 m.

What is its speed after accelerating?

Answer 1

Speed is 34.5 m/s

Step-by-step explanation:

From the third equation of motion:

`{ \boxed{ \sf{v {}^(2) = {u}^(2) + 2as}}}`

v is final velocity, u is initial velocity, a is acceleration, s is displacement.

substitute:

`{v}^(2) = {30}^(2) + (2 * 2.4 * 60) \\ {v}^(2) = 900 + 288 \\ {v} {}^(2) = 1188 \\ {v} = √(1188) \\ v = 34.5 \: m {s}^( - 1)`