Posted by Dana on Saturday, September 4, 2010 at 5:23am.
a car starting from rest is accelerated at 2m/s(square) for 10s, then moves at constant velocity for 20s and then decelerates at 1m/s(square), finally stops. how far does it travel during its trip?
please help me! I need in solution for that problem

physics  Damon, Saturday, September 4, 2010 at 5:55am
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

physics  help me, Wednesday, May 15, 2013 at 6:05pm
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?

physics  Anonymous, Saturday, September 10, 2016 at 4:09am
(a)80m
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