Log in
Physics
1
answer
0
watching
129
views

The position-time equation for a certain train is xf=2.8m+(8.8m/s)t+(2.3m/s2)t2. What is the acceleration?
I tried 2.3m/s2 but it was wrong

Watch
1
answer
0
watching
129
views

For unlimited access to Homework Help, a Homework+ subscription is required.

Unlock all answers

Get 1 free homework help answer.
Already have an account? Log in
Share

Related textbook solutions

Physics: Principles with Applications
7th Edition, 2013
Giancoli
ISBN: 9780321625922

Related questions

At time t1 = 2.00 s, the acceleration of a particle in counterclockwise circular motion is (7.00 m/s2) ihat + (8.00 m/s2) jhat. It moves at constant speed. At time t2 = 5.00 s (3/4 of a revolution later), its acceleration is (8.00 m/s2) ihat + (-7.00 m/s2) jhat. What is the radius of the path taken by the particle?



(a) If the maximum acceleration that is tolerable for passengers ina subway train is 1.44 m/s2 and subway stations are located 826 mapart, what is the maximum speed a subway train can attain betweenstations? (b) What is the travel time between stations? (c) If asubway train stops for 24.0 s at each station, what is the maximumaverage speed of the train, from one start-up to the next?


Weekly leaderboard

Home
Homework Help 3,900,000
Physics 110,000
Start filling in the gaps now
Log in