a vehicle can accelerate from 0 to 60 mi/h (26.8 m/s) in 4.0 s. What is the average acceleration during this time interval?
2006-07-13 05:36:53 · 8 answers · asked by Lou 1 in Science & Mathematics ➔ Physics
average accelration = change in velocity/total time taken
= (26.8-0)/(4/0)
= 6.7 m/s^2
It's around two-thirds of acceleration due to gravity.
2006-07-13 05:43:38 · answer #1 · answered by haroun i 2 · 0⤊ 0⤋
Average acceleration = (v1-v2)/(t2-t1) or in this case:
(60 - 0)/(4 - 0) = 15 miles/hour/second.
Using the 26.8 m/s, you get 26.8/4 = 6.7 m/s/s or m/s^2
2006-07-13 06:14:17 · answer #2 · answered by tbolling2 4 · 0⤊ 0⤋
Average acceleration= change in speed/ time
= (26.8-o)/4
=6.70 ms-2
2006-07-13 05:51:23 · answer #3 · answered by léa 2 · 0⤊ 0⤋
average acceleration = (change in velocity) / (change in time) Therefore: Average acceleration = (11m/s - 5 m/s )/ 4s Answer: (6m/s) / 4s = 1.5 m/s/s = 1.5 m/s^2
2016-03-27 03:52:13 · answer #4 · answered by Anonymous · 0⤊ 0⤋
final speed= intial speed + acceration x time
intial speed=0 m/s,, final speed = 26.8m/s,, time=4 sec
put the value in the above relation ,,,,
answer= 6.7 meter/second square
2006-07-13 06:34:40 · answer #5 · answered by Atul 1 · 0⤊ 0⤋
Go Steelers
2006-07-13 05:40:20 · answer #6 · answered by nflramey 1 · 0⤊ 0⤋
26.8/4 m/s^2 = 6.7m/s^2
2006-07-13 05:39:37 · answer #7 · answered by ag_iitkgp 7 · 0⤊ 0⤋
26.8/4=6.7 m/s*s
2006-07-13 05:54:36 · answer #8 · answered by Anonymous · 0⤊ 0⤋