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alexira [117]
3 years ago
9

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.
Physics
2 answers:
melamori03 [73]3 years ago
8 0
<u>Answer</u>

66m

<u>Explanation</u>
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

zubka84 [21]3 years ago
5 0
<span>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 </span>
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A man starts walking from home and walks 2 miles at 20° north of west, then 4 miles at 10° west of south, then 3 miles at 15° no
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Answer:

a)  R = 2.5 mi   b)  To return to your case you must walk in the opposite direction or θ = 98º

This is 8º north west

Explanation:

This is a distance exercise with vectors the best way to work these is to decompose the vectors and perform the sum on each axis separately

To use the Cartesian system all angles must be measured from the positive side of the x-axis or the signs of the components must be assigned manually depending on the quadrant where they are.

First vector A = 2 to 20º north west

Measured from the positive x axis is θ = 180 -20 = 160º

We use trigonometry to find the components

     Cos 20 = Aₓ / A

     sin 20 = A_{y} / A

    Aₓ = A cos 160 = 2 cos 160

    A_{y}  = A sin160 = 2 sin160

    Aₓ = -1,879 mi

    A_{y}  = 0.684 mi

Second vector B = 4 mi 10º west of the south

Angle θ = 270 - 10 = 260º

    cos 2600 = Bₓ / B

    sin 260 = B_{y} / B

    Bₓ = B cos 260

     B_{y}  = B sin 260

    Bₓ = 4 cos 260

     B_{y}  = 4 sin 260

     Bₓ = -0.6946mi

     B_{y}  = - 3,939 mi

Third vector C = 3 mi to 15 north east

     cos 15 = Cₓ / C

     sin15 = C_{y} / C

     Cₓ = C cos 15

     C_{y} = C sin15

     Cₓ = 3 cos 15

    C_{y} = 3 sin 15

     Cₓ = 2,898 mi

    C_{y} = 0.7765 mi

Now we can find the final position of the person

    X = Aₓ + Bₓ + Cₓ

    X = -1.879 -0.6949 + 2.898

    X = 0.3241 mi

    Y = A_{y} +  B_{y} + C_{y}

    Y = 0.684 - 3.939 +0.7765

    Y = -2.4785 mi

a) We use Pythagoras' theorem

     R = √ (x2 + y2)

     R = √ (0.3241 2 + (-2.4785) 2)

     R = 2.4996 mi

     R = 2.5 mi

b) let's use trigonometry

     Tan θ = y / x

     Tanθ  = -2.4785 / 0.3241

     θ = tan⁻¹ (-7,647)

     θ = -82

Measured from the positive side of the x axis is Te = 360 - 82 = 278º

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To return to your case you must walk in the opposite direction or Te = 98º

This is 8º north west

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That is;

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Where;

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F = force (N)

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Based on the information provided in this question, F = 5000N, d = 0.25m

Hence;

W = F × d

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