A car starts from rest at a stop sign. It accelerates at 4.0 m/s2 for 3 seconds, coasts for 4 s, and then slows down at a rate o

Question

sergiy2304 [10]

3 years ago

11

A car starts from rest at a stop sign. It accelerates at 4.0 m/s2 for 3 seconds, coasts for 4 s, and then slows down at a rate o

f 3.0 m/s2 for the next stop sign. How far apart are the stop signs?

Physics

2 answers:

solmaris [256] · Answer 1 · 2022-07-16T20:14:09+03:00

Well, Break the problem up into parts. For speeding up: The car accelerates at 4 m/s^2 for 3 seconds. Multiplying these values tells you the car reaches a speed of 12 m/s. Vf^2 = Vi^2 + 2a(Xf - Xi) 12^2 = 0 + 2(4)(Xf - 0) 144 = 8 Xf Xf = 18 m For coasting: The car is at the 12 m/s and does this for 2 seconds. x = vt = (12)(2) = 24 m For slowing down: The car decelerates at 3 m/s^2 to come to a stop at the next sign. From 12 m/s, this would take 12/3 seconds or 4 s. Vf^2 = Vi^2 + 2a(Xf - Xi) 0 = 12^2 + (2)(-3)(Xf - 0) -144 = -6 Xf Xf = 24 m Summing the distances: Xtotal = 18 + 24 + 24 = 66 m

Nataly_w [17] · Answer 2 · 2022-07-16T22:18:35+03:00

Nataly_w [17]3 years ago

3 0

Wow um what @unbroken said