Posted by Dana on Saturday, September 4, 2010 at 5:23am.
Stage 1, acceleration
a = 2 m/s^2
v = 0 + 2 * 10 = 20m/s
x = 0 + 0*10 + .5 (2)(10^2) = 100 m
Stage 2, starting with v = 20 and x = 100
a = 0
v = 20 + 0 * 20 = 20
x = 100 + 20(20) +.5*0*20^2 = 500
stage 3, starting with v = 20 and x = 500
a = -1
v = 20 -1(t) = 20 -t
v = 0 at end so t = 20 for deacceleration
x = 500 + 20(20) -.5(20)^2
= 500 + 400 - 200
= 700
For this problem:
Change in speed over change in time is acceleration...
A car travels on a straight, level road. Starting from rest, the car is going 20 m/s at the end of 2.0 sec.
(a) What is the acceleration of the car during the first 2.0 seconds of motion?
In 7.0 more seconds, the car is going 40 m/s.
(b) What is the car's acceleration for this period of time (during the 7 seconds)
The car then slows to 19 m/s in 2.0 seconds.
(c) What is the acceleration of the car during this period of time?
(d) What is the overall average acceleration of the car for the total time?
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