Question

A car starts from rest at a stop sign. It accelerates at 4.0 m/s^2 for 3 seconds, coasts for 2 s, and then slows down at a rate of 3.0 m/s^2 for the next stop sign. Determine the distance between the two stop signs.

Answer 1

Answer

66m

Explanation
This question will involve the three linear equations
1st v=at+u
2nd s=ut+1/2at²
3rd v²=u²+2as

Distance for part 1
s=ut+1/2at²
  =0×3 + 1/2×4×3²
  = 18m

Distance for part 2
v=at+u
  =4×3 +0
  =12
distance = speed × time
              =12×2
              = 24m
Distance for part 3
v²=u²+2as
0=12²-2×3×s
144 = 6s

s = 144/6
   = 24

total distance = 18+24+24
                      = 66m

Answer 2

FIRST SECTION You should use the formula for uniformly accelerated linear movement. Initial speed is 0 because it starts from rest. d=(1/2)*a*t^2+vo*t =(1/2)*(4.0 m/s^2)*(3s)^2+0*3s=(1/2)*(4.0 m/s^2)*3^2*s^2+0=2.0 m*9=18m You can calculate the final speed with the other formula: v=a*t+vo=(4.0 m/s^2)*(3s)+0=(4.0 m/s)*(3)=12m/s SECOND SECTION You should use the formula for uniform linear movement. Velocity is a constant: it remains in 12m/s. d=v*t=12m/s*2s=12m*2=24m THIRD SECTION We should use the same formulas as the first section, but with different numbers. Initial velocity will be 12m/s, and then velocity will start to decrease until it gets to 0. We don’t know what the time is for this section. Acceleration is negative, because it’s slowing down. v=a*t+vo 0=-3.0 m/s^2*t+12m/s 3.0 m/s^2*t=12m/s t=(12m/s)/(3.0 m/s^2)=4(1/s)/(1/s^2)=4s^2/s=4s Now let’s use that time in the other formula: d=(1/2)*a*t^2+vo*t =(1/2)*(-3.0 m/s^2)*(4s)^2+(12m/s)*3s=(-1.5 m/s^2)*4^2*s^2+12*3m*s/s=-1.5 m*4^2+36m=-1.5*16m+36m=-24m+36m=12m Now let’s add the 3 stages: d=18m+24m+12m=54m