Velocity question two parts

In summary, Sue is driving at 35.0 m/s and encounters a slow-moving van 155 m ahead traveling at 5.30 m/s. With an acceleration of -2.00 m/s2, Sue attempts to apply her brakes. Using the formula s = ut + 1/2at^2, it is determined that the collision will occur at 6.76 seconds and 191 meters into the tunnel.
  • #1
Melchior25
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0

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.
 
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  • #2
6.755 s; 191 m?
 
  • #3
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?
 
  • #4
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:
[tex] s = ut +\frac{1}{2}at^2[/tex]
[tex] s_{sue} = 35t + \frac{1}{2}(-2)t^2[/tex]

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

[tex] s_{sue} = s_{van} [/tex]

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 [tex]s_{sue} \ or \ s_{van}[/tex], you get 191 m.
 

1. What is velocity?

Velocity is a measure of how fast an object is moving in a specific direction. It is a vector quantity, meaning it has both magnitude (speed) and direction.

2. How is velocity different from speed?

While speed only measures how fast an object is moving, velocity also takes into account the direction in which the object is moving. This means that two objects can have the same speed but different velocities if they are moving in different directions.

3. How do you calculate velocity?

Velocity is calculated by dividing the distance an object has traveled by the time it took to travel that distance. This is represented by the formula v = d/t, where v is velocity, d is distance, and t is time.

4. What are the units of velocity?

The units of velocity are typically expressed as distance over time, such as meters per second (m/s) or kilometers per hour (km/h). However, they can also be written in other units such as miles per hour (mph) or feet per second (ft/s).

5. How does velocity relate to acceleration?

Velocity and acceleration are closely related, as acceleration is the rate of change of velocity over time. This means that an object with a constant velocity has no acceleration, while an object with a changing velocity has a non-zero acceleration. The formula for acceleration is a = (v2 - v1)/t, where a is acceleration, v2 is the final velocity, v1 is the initial velocity, and t is the time interval.

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