A teenager has a car that speeds up at 3.00 m/s2 and slows down at -4.50 m/s2. O

Question

A teenager has a car that speeds up at 3.00 m/s2 and slows down at -4.50 m/s2. On a trip to the store, he accelerates from rest to 12.0 m/s, drives at a constant speed for 5.00 s, and then comes to a momentary stop at an intersection. He then accelerates to 18.0 m/s, drives at a constant speed for 20.0 s, slows down for 2.67 s, continues for 4.00 s at this speed, and then comes to a stop. (a) How long does the trip take? (b) How far has he traveled? (c) What is his average speed for the trip? (d) How long would it take to walk to the store and back if he walks at 1.50 m/s?

Explanation / Answer

(A) vf = v0 + a t

d = v0 t + a t^2 /2

and vf^2 - vi^2 =2 a d

0 to 12 m/s :

t1 = 12/3 = 4 sec  

12^2 - 0^2 = 2( 3)d1

d1 = 24 m

constant 12 m/s:

t2 = 5 sec  

d2 = 12 x 5 = 60 m

12 m/s to 0 m/s :

t3 = (0 - 12)/4.5 = 8/3 sec  

d3 = 12^2 / (2 x 4.5) = 16 m

0 to 18 m/s :

t4 = 18/3 = 6 sec

d4 = 18^2 / (2 x3) = 54 m

constant speed :

t5 = 20 s

d5 = 20 x 18 = 360 m

slow down for 2.67 sec :

t6 = 2.67 sec  

d6 = (18 x 2.67) - (4.50 x 2.67^2 / 2)

d6 = 32 m

constant :

v = 18 - (4.50 x 2.67) = 6 m/s

t7 = 4 sec  

d7 = 24 m

(A) t = t1 + t2 + t3 + t4 +t5 + t6+ t7

t = 44.3 sec  

(b) d = d1 + d2 +d3 + d4 + d5 + d6 + d7

d = 570 m

(C) <v> = d / t = 12.9 m/s

(d) t = 570 / 1.50 = 380 sec Or 6.33 min
  

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