55.8 ft/sec - 45 ft/sec = 10.8 ft/sec
10.8ft/sec/6.30ft/sec^2 = 1.714 sec. = 1.7 seconds
d) determine how much of the 300 ft displacement is due to the initial velocity and how much is due to the acceleration
e) find the time required for traveling the first 150 ft.
(76ft/sec + 45 ft/sec.)/2 = 121/2 = 60.5 ft/ sec.
300 ft / 60.5 ft/ sec. = 4.9 seconds??
acceleration = m = a = change in velocity divided by change in time = (76ft/sec - 45 ft/sec)/4.9 sec. = 31ft/sec/4.9 sec. = 6.3 ft/sec^2
c) find the time when the speed of the car will be 55.8 ft/sec. ...
Initially you are driving at 45 ft/sec on a straight road. You accelerate the car at a steady rate and increase the speed to 76ft/sec. During this acceleration process you travel a distance of 300 ft.
a) Calculate the time required for the 300 ft travel...
But the instantaneous speed appears to be the same number as the average speed, which is 22.5 mi/hr. Now what about: what about the time when the speedometer will read the average speed you calculated for part a) as the instantaneous speed??
Show that Newton's Second Law is NOT valid in a reference frame moving past the laboratory frame of problem 1 with a constant acceleration?
Problem 1: In a laboratory frame of reference, an observer notes that Newton's Second Law is valid. Show that it is also valid for an...
9 mi/hr/sec. times 2.5 seconds = 22.5 mi/hr because the seconds cancel??
So I get the instantaneous from multiplying the acceleration times the time? which would be 9 mi/hr/sec. times 5 seconds = 45 mi/hr?? I am confused.
At what distance (OR at what time) does it take for an object falling at 32 ft per sec per sec. to reach terminal velocity in its descent??
The only equation I can think of to use here would be distance = (rate of descent) times time.
The Attempt at a...
So, for c): the speed of the car at t = 2.5 seconds = velocity = accel. times the time = 9 mi/hr/sec. times 2.5 seconds = 22.5 mi/hr/sec^2 and for d) it's the same answer??
How does speed equal the velocity which equals the acceleration times the time?? I don't get that.
a) the average speed during the 5 seconds
b) the acceleration in miles per second at t = 3.2 sec. after you start your car = 9 mi/hr/sec??
c) the speed of the car at t = 2.5 sec.
d) the time when the speedometer will read the average speed you calculated for part a) as the instantaneous speed.