Will Speedy Sue Collide with the Van in the Tunnel?

[Melchior25]

Homework Statement

Speedy Sue, driving at 35.0 m/s, enters a one-lane tunnel. She then observes a slow-moving van 155 m ahead traveling at 5.30 m/s. Sue applies her brakes but can accelerate only at -2.00 m/s2 because the road is wet. Will there be a collision?
yes
If yes, determine how far into the tunnel and at what time the collision occurs.

Homework Equations

0=at^2-bt+c
where a is the given acceleration, b is the combined velocities, and c is that specified initial distance that the cars were from each other.

Xf = Xi + (Vi)t +0.5(a)(t^2) for Sue's car and set it equal to Xf = Xi + (Vi)t

155+5.3*t= distance (m)

The Attempt at a Solution

I ran the numbers through the quadratic equation and got 23.4543 s for my time. I then plugged that time value into the distance equation above and got 279.308 m. However both of my answers are wrong. I'm a bit lost and could use a little help.

 

[Leong]

6.755 s; 191 m?

 

[Melchior25]

Thanks Leong, I have not put the answers in yet to see if they are correct because the site is down. I just wanted to know how you got your solutions, because I'm just curious as to where I went wrong. Such as what equations did you use?

 

[Leong]

our reference point / origin of our displacement will be that of sue when she enters the tunnel. so all the followed displacement will refer to that point.
Direction to the right is treated as positive.

Formula used:
$$s = ut +\frac{1}{2}at^2$$
$$s_{sue} = 35t + \frac{1}{2}(-2)t^2$$

$$s_{van} = 155 + 5.3t$$ [when sue enters the tunnel, the van is already 155 m ahead of her]
if collision happens,

$$s_{sue} = s_{van}$$

solve for t, you get 22.9 s and 6.76 s.
first answer is omitted because sue will have stopped by 17.5 s.
substitute t = 6.76 s into $$s_{sue} \ or \ s_{van}$$, you get 191 m.