The equation that solves a problem is . The problem is:?

Question

The equation that solves a problem is (18m/s)^2 - (0m.s)^2 = 2(3.0m/s^2)(3.0m) . The problem is:

What is the initial velocity of a car that goes from rest to 18 m/s in 3.0 s?

What is the final velocity of a car that goes from rest to 18 m/s in 3.0 s?

What is the initial velocity of a car that accelerates at 18 m/s for 3.0 s?

What is the final velocity of a car that accelerates at 3.0 m/s2 over a 6.0 m distance?

What is the final velocity of a car that accelerates at 3.0 m/s2 over a 3.0 m distance?

Answer 1

hi,

(18m/s)^2 - (0m.s)^2 = 2(3.0m/s^2)(3.0m)

=> vf^2 - vi^2 = 2aS

What is the initial velocity of a car that goes from rest to 18 m/s in 3.0 s? 0 m/s as it was initially at rest

What is the final velocity of a car that goes from rest to 18 m/s in 3.0 s? 18 m/s as it reaches to this value after 3.0 s

What is the initial velocity of a car that accelerates at 18 m/s for 3.0 s? you have used wrong units for acceleration, it shall be m/s^2, however, vf = vi + at, assuming vi = 0

=> vi = vf - at, you also have to give the final velocity. recheck your statement.

What is the final velocity of a car that accelerates at 3.0 m/s2 over a 6.0 m distance? assuming the initial velocity was 0 m/s

2aS = vf^2 - vi^2 => 2(3)(6) = vf^2 - 0, => 36 = vf^2 => vf = 6 m/s

What is the final velocity of a car that accelerates at 3.0 m/s2 over a 3.0 m distance?  assuming the initial velocity was 0 m/s

2aS = vf^2 - vi^2 => 2(3)(3) = vf^2 - 0, => 18 = vf^2 => vf = 3sqrt(2) m/s

hope to answer you well. bye.

Answer 2

from the equation

v = 18

u = 0

a = 3

s = 3

The equation is of the form:

v^2 - u^2 = 2as

use these parameters and solve using the standard equations putting question parameters as well...