Question

A car traveling at 42 ft/sec decelerates at a constant 2 feet per second squared. How many feet does the car travel before coming to a complete stop?

Answer 1

use one of equations of constant acceleration
u = 42 , v = 0 , a = -2, s = ?

v^2 = u^2 + 2 as
0 = 42^2 - 2 * 2 * s
4s = 42^2

s = 42 * 10.5 = 441 feet

Answer 2

that is not my answer right?

Answer 3

u = initial velocity, v = final velocity, a = acceleration and s = distance

Answer 4

yes thats the answer - note a = -2 because the car is decelerating

Answer 5

okay. the -2 accerlration was what was getting me.. thanks. can i ask another question for you?

Answer 6

ok

Answer 7

so i can get everything, but the last line is not working for me.

Answer 8

sorry - i can't help you with that - i've never come across riemann sums before

Answer 9

its okay thanks for your help though

Answer 10

yw