AP Physics Question, Again?

Question

A car accelerates from rest at 4.2 m/s2 for 4 s, travels at a constant speed for 7 s, accelerates at 1.0 m/s2 for 18 s, and then decelerates to rest at 2.9 m/s2 (Fig. 2-30, not to scale). How far has the car traveled?

I have no idea how to solve this.

Answer 1

I think all you have to do is work every problem out....so, 4.2 times the 4 seconds, plus 4.2 times 7, plus 1 times the 18 seconds, and minus whatever the distance is times 2.9.  I'm guessing the fig. 2-30 is an illustration so just ignore that.

16.8m

29.4m

18m(add all these)

64.2m

^...this answer is then divided by 2.9

and add the answer to 64.2

I hope this helps...or that it's right!?

Oh and i saw the other answer, the car couldn't have traveled only 14.blah blah meters so check out what she did cause yeah....

Answer 2

4.2x4=16.8

16.8+7=23.8

23.8+18= 41.8

41.8 divided by 2.9= 14.413

Final Awnser: 14.413

Answer 3

Take them one part at a time

accelerates from rest at 4.2 m/s2 for 4 s

d = ½at²  = ½(4.2)4² = 33.6 m

v = at = 4.2*33.6 = 141.1 m/s

travels at a constant speed for 7 s

d = vt = 141.1 x 7 = 987.8 m

accelerates at 1.0 m/s2 for 18 s

∆V = at = 1*18 = 18 m/s

d = ½at² + v₀t = ½1*18² + 141.1*18

= 162 + 2539.8 = 2701.8 m

v = ∆V + v = 18 + 141.1 = 159.1 m/s

then decelerates to rest at 2.9 m/s2

same as acc, with opposite sign

v = at

159.1 = 2.9t

t = 54.9 sec

d = ½at² = ½(2.9)(54.9)² = 4370 m

total is

33.6 + 987.8 + 2701.8 + 4370 = 8094 m

if I didn't make an arith mistake.

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