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?

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