Need help to sole this kinetic homework?

Car A is 600 m from the centre of the intersection when it starts from rest and accelerates uniformly at 4m/ss. Reaching a speed 0f 24m/s that it maintains. Art the instant car A take off, car B IS 780 m from the centre of the intersection and travelling at a constant speed of 28m/s. When car B is 53 m from the centre of the intersection it decelerates uniformly at 5m/ss.
a. Which car gets to the centre of the intersection first?
b. How far past the centre of the intersection is the first car when the second car reaches it?
c. If all other condition remains the same, what constant acceleration would:
- The second car needs to have for a collision to occur?
- The first car needs to have for a collision to occur?



I have done the question a and b …but I don’t know how to do question c. Anyone help me?

the result for a is
a. t (A)= 28 second.
t(B)= 28.35 second.

so car A comes first.

b. it's 8 m from the intersection to the first car.

c. ???? i know the answer but i don't know how to work out ...it's
- a= -2m/ss
- a= 3.58m/ss
Can anyone help me with question c? please ...

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.

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

--------
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 ------******

------
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.

------------------
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.

--------------------
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.

-------
- 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.

It's absolutely great ...thank you very much for helping me to solve this exercise in very detailed ....thanks ....