The answers to part (c.) are dependent on the answers in part (a.). Or, the answers of (c.) are relative to the answers of (a.)---specifically on the precision of your t(B).

Your numbers here are in two decimal places, so let us use two decimal places for all answers.

In this case, your t(B) = 28.35 sec is not precise. The t(B) must be 28.37 sec.

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a. Which car gets to the centre of the intersection first?

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For car A:

Time from 0 to 24 m/sec,

V = Vo*t +a*t

24 = 0*t1 +4*t1

t1 = 24/4 = 6 sec ------***

Distance traveled in that 6-sec acceleration,

X = Vo*t +(1/2)a*t^2

X = 0*6 +(1/2)(4)(6^2) = 72 meters.

Time for the rest of the 600 m,

X = v*t

(600 -72) = 24*t2

t2 = 528/24 = 22 sec -----***

Therefore, for total time,

t(A) = t1 +t2 = 6 +22 = 28 sec ------******

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For car B:

t(B) = (780 -53)/28 + (time spent in those 53 meter)

t(B) = 25.96 +say, (tb) ------(i)

X = Vo*t +(1/2)a*t^2

53 = 28*tb +(1/2)(-5)(tb)^2

53 = 28(tb) -2.5(tb)^2

2.5(tb)^2 -28(tb) +53 = 0

Using the Quadratic Formula,

tb = {-(-28) +,-sqrt[(-28)^2 -4(2.5)(53)]} / (2*2.5)

tb = (28 +,-15.94)/5

tb = 8.79 sec, or 2.41 sec

If tb = 8.79 sec,

V = Vo +a*t

V at center of intersection = 28 +(-5)(8.79) = -15.95 m/sec.-----cannot be, so reject tb=8.79 sec.

Therefore, tb = 2.41 sec

Substitute that into (i),

t(B) = 25.96 +2.41 = 28.37 sec ------------********

So, t(A) is less than t(B).

Therefore, car A gets first to the center. --------answer.

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b. How far past the centre of the intersection is the first car when the second car reaches it?

1st car to reach the center is car A, so we use v = 24 m/sec.

time t = 28.37 -28 = 0.37 sec

X = V*t = 24(0.37) = 8.88 meters ----answer.

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c. If all other condition remains the same, what constant acceleration would:

- The second car needs to have for a collision to occur?

This means the two cars reach the center of the intersection at the same time as t(A).

Or, t(A) = t(B) = 28 sec

So, in (i),

t(B) = 25.96 +tb = t(A) = 28

25.96 +tb = 28

tb = 28 -25.96 = 2.04 sec

Then,

53 = 28(2.04) +(1/2)a(2.04)^2

53 = 57.12 +2.08a

a = (53 -57.12)/2.08 = -1.98 m/sec/sec -----answer.

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- The first car needs to have for a collision to occur

Here it means the two cars reach the center of the intersection at the same time as t(B).

Or, t(A) = t(B) = 28.37 sec

a) For the acceleration interval of car A:

V = Vo +a*t

24 = 0 +a*t1

t1 = 24/a sec ----***

X = Vo*t +(1/2)a*t^2

X = 0*t +(1/2)a(24/a)^2

X = 288/a meters

b) For the rest of the 600m:

distance = Velocity*time

(600 -X) = 24*t2

t2 = (600 -X)/24

t2 = (600 -288/a)/24

t2 = (25 -12/a) sec -----***

So,

Total t(A) = t1 +t2 = t(B) = 28.37

24/a +(25 -12/a) = 28.37

24/a -12/a = 28.37 -25

12/a = 3.37

a = 12/3.37 = 3.56 m/sec/sec --------answer.