Velocity and Distance Calculations for a Cyclist Riding Down a Hill

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In summary: And you have to find out t. :wink:and for e) both drivers suddenly see a barrier 100m away and hit their brakes at the same time. assuming both vehicles decelerate uniformly, and they take 3 seconds to stop, will they stop in time?:yuck: ~Amyfor e) i calculated the acceleration at 8.3 m/s for the truck, and 34.33 m/s for the car. and for both used the formula:d = (v)(t) + (0.5)(a)(t^2).for the truck i got 112.48 m and for the car i got 463.5 m i
  • #1
physicsgal
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i have the answers.. but just want to make sure I am on the right track and didnt mix anything up.

"a cyclist rode east for 100m with a constant velocity of 25m/s. she then accelerated down a hill and 5 seconds later reached the bottom of the hill with a velocity of 50m/s."

a) how long did the cyclist ride before reaching the hill?

100/25 = 4 seconds

b) what was her average acceleration down the hill?
(50m/s -25.0 m/s)/5s
= 5 m/s ^2

*not sure if the 25 m/s should have been left out of the answer or not*

c) what was the distance she traveled down the hill?
5s * 50m/s
= 250 m

d) if she continued to ride at constant velocity for another 5s, how far beyond the hill would she travel?
5m * 25 m/s
= 125m

any help will be appreciated!

~Amy
 
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  • #2
a,b - correct
c,d - wrong
Gotta go, I have class now.
 
  • #3
(a) and (b) are correct. :wink:

(c). To find the distance traveled down the hill you should use the following equation:
distance = average velocity x time

(d) displacement = velocity x time
Why did you use 25 m/s for velocity if the cyclist reaches the bottom of the hill with a velocity of 50 m/s? :grumpy:
 
  • #4
thanks guys!

for c) i now have:
0.5(50+25)(5s)
=187.5 m

or should i also include the 4 seconds that i calculated in a)?

for d) i used the 25m/s because in the question it was called the "constant velocity". :cry:

so for d) should i use 0.5(50+25) (the average) or just 50m/s for the velocity? since she's traveling "beyong the hill" (straight probably - not downhill) she's probably moving slower than 50 m/s as she was going down the hill. imo, she'd be going the same velocity as she went she was just biking TO the hill which was 25m/s.

correct me if I am wrong. :tongue:

~Amy
 
  • #5
physicsgal said:
thanks guys!

for c) i now have:
0.5(50+25)(5s)
=187.5 m

or should i also include the 4 seconds that i calculated in a)?
Your answer is correct. You don't include the 4 second calculated in a) because you are calculating the distance she traveled down the hill so the time calculated in a) as nothing to do with it.

physicsgal said:
for d) i used the 25m/s because in the question it was called the "constant velocity". :cry:
The velocity is constant only before the cyclist went down the hill. When he is traveling down the hill, he has acceleration. Thus, the velocity cannot be constant.

physicsgal said:
so for d) should i use 0.5(50+25) (the average) or just 50m/s for the velocity? since she's traveling "beyong the hill" (straight probably - not downhill) she's probably moving slower than 50 m/s as she was going down the hill. imo, she'd be going the same velocity as she went she was just biking TO the hill which was 25m/s.
You use 50 m/s for the velocity, which is both the instantaneous and average velocity, because, after she reached the bottom of the hill, it is constant.

I can see that you are messing things a little. Be careful with that, please. :smile:
 
  • #6
thanks for the tips! :biggrin:

another one:

if a car is accelerating 4 m/s ^2, how far of distance does it travel after 10 seconds?

the formula I am using is (0.5)(4^2)(10)^2 = 800 m.

is that right? :shy: or should i leave out the 0.5 so it'd equal 1600 m?

~Amy
 
  • #7
physicsgal said:
thanks for the tips! :biggrin:
My pleasure. :smile:

physicsgal said:
another one:

if a car is accelerating 4 m/s ^2, how far of distance does it travel after 10 seconds?

the formula I am using is (0.5)(4^2)(10)^2 = 800 m.
~Amy
You may use the following equation:
[tex]x(t) = v_0t + \frac{1}{2}at^2[/tex]

It appears that:
v0= 0 m/s
a = 4 m/s2

physicsgal said:
is that right? :shy: or should i leave out the 0.5 so it'd equal 1600 m?

~Amy
Why should you leave out 0.5? Why should it be equal to 1600 m? :shy:
 
  • #8
thanks again! i was using the formula you mentioned, but don't understand why the 0.5 is there. :blushing:

im finally on the right track :smile:

more to the question:

"the car starts from rest and accelerates uniformly at 4 m/s east. at the same instant, a truck traveling with a constant velocity of 90 km/hr east overtakes and passes the car".

c) the car passes the truck at a distance of 312.5 m beyond the starting point. how fast is the car traveling at that instant?


so the truck is traveling at 25 m/s (according to my calculation)

would the formula be:
(25m^2) + 2(4^2)(312.5)
= 10625 square rooted
= 103.08 m/s
= 10.15 m/s ^2?

~Amy
 
Last edited:
  • #9
for d) the question is how long does the car take to pass the truck?

4.8 seconds?

and for e) both drivers suddenly see a barrier 100m away and hit their brakes at the same time. assuming both vehicles decelerate uniformly, and they take 3 seconds to stop, will they stop in time?

:yuck:

~Amy

for e) i calculated the acceleration at 8.3 m/s for the truck, and 34.33 m/s for the car.

and for both used the formula:
d = (v)(t) + (0.5)(a)(t^2).

for the truck i got 112.48 m

and for the car i got 463.5 m

i dunno.. any help will be appreciated

~Amy
 
Last edited:
  • #10
physicsgal said:
thanks again! i was using the formula you mentioned, but don't understand why the 0.5 is there. :blushing:
Do you know how to find/work out that formula?

physicsgal said:
"the car starts from rest and accelerates uniformly at 4 m/s east. at the same instant, a truck traveling with a constant velocity of 90 km/hr east overtakes and passes the car".

c) the car passes the truck at a distance of 312.5 m beyond the starting point. how fast is the car traveling at that instant?


so the truck is traveling at 25 m/s (according to my calculation)

would the formula be:
(25m^2) + 2(4^2)(312.5)
= 10625 square rooted
= 103.08 m/s
= 10.15 m/s ^2?

~Amy
You correctly used the formula [itex]v_f^2 = v_i^2 + 2a \Delta x[/itex] :approve:. But you made some mistakes when calculating. Would you look at that again, please? :smile:
 
  • #11
physicsgal said:
for d) the question is how long does the car take to pass the truck?

4.8 seconds?
Unfortunetely, no.
This formula may be useful:
[tex]x(t) = (1/2)at^2[/tex]

You know the values of:
x(t) and a
Thus, you can find t.

physicsgal said:
for e) i calculated the acceleration at 8.3 m/s for the truck, and 34.33 m/s for the car.
~Amy
How did you calculated those accelerations? :rolleyes:
Also, be careful with units. Acceleration is expressed in m/s2 (SI units).
 
  • #12
You correctly used the formula . But you made some mistakes when calculating. Would you look at that again, please?

i checked it again but got the same answer :yuck:

Originally Posted by physicsgal
for e) i calculated the acceleration at 8.3 m/s for the truck, and 34.33 m/s for the car.
~Amy

How did you calculated those accelerations?
25/3 = 8.333 (truck)
103/3 = 34.33 (car)

Originally Posted by physicsgal
thanks again! i was using the formula you mentioned, but don't understand why the 0.5 is there.


Do you know how to find/work out that formula?
im going to look into it. but I am not very mathimatical.


Unfortunetely, no.
This formula may be useful:


You know the values of:
x(t) and a
Thus, you can find t.

what variable is x(t)?
just to be clear:
the "a" = acceleration
"t" = time in seconds?

thanks for the help! i really appreciate it :smile:

~Amy
 
  • #13
update!

for c) i figured it out the old fashion way (creating a graph) to find the time for 312.5 (based on acceleration of 4 m/s^2) and ended up with 50 m/s. i hope that's right!

~Amy
 
  • #14
im still having trouble with c) :confused:
my current calculation is (4.0^2)(12.5) = 200 m/s.. but that doesn't sound right or is it?

and for d) i got 12.5 seconds

~Amy
 
  • #15
physicsgal said:
thanks again! i was using the formula you mentioned, but don't understand why the 0.5 is there. :blushing:

~Amy

The .5 is there because of the relationship between acceleration, velocity, and position.

velocity is the derivitive of position.

x = x0 + v0t + 1/2at^2
x is position
v = v0 + at
v is velocity

so the 1/2 is there because t is squared, and it's needed to offset the 2 when you diferentiate.

might not be too clear, but I hope that helped
 
  • #16
thanks bigcheese!

and for e) both drivers suddenly see a barrier 100m away and hit their brakes at the same time. assuming both vehicles decelerate uniformly, and they take 3 seconds to stop, will they stop in time?

does anyone have the foggiest idea how to solve this one?
without knowing the acceleration rates, is it even possible?

and for c) my latest calculation is 100m/s.. it really depends what formula i use :uhh:

~Amy
 
  • #17
and for e) both drivers suddenly see a barrier 100m away and hit their brakes at the same time. assuming both vehicles decelerate uniformly, and they take 3 seconds to stop, will they stop in time?

in the equation

x = x0 + v0t + .5at^2

remember that a is going to be a negative value, because they're decelerating. See if that helps at all.
 
  • #18
ah, also, this just occurred to me. if they go from v to 0 uniformly, and it takes them 3 seconds to do that, all you need is v to determine their accel.

a = change in velocity, so if they're going 90, and in 3 secs they go 0, their accel would be -30m/s^2
 
  • #19
update:

for a) i got 200 m/s (my original answer of 800 m was wrong)
c) 50 m/s
d) 12.5s
e) 37.65 m for the truck, and 75 m for the car. for deceleration i got -8.3m/s^2 for the truck, and -16.67 m/s^2 for the car.

:smile:

thanks thebigcheeze and PPonte!

~Amy
 

1. What is the difference between velocity and speed?

Velocity is a vector quantity that includes both speed and direction, while speed is a scalar quantity that only measures the rate of motion.

2. How is velocity calculated?

Velocity is calculated by dividing the change in distance by the change in time. This can be represented by the equation velocity = distance/time.

3. What is the relation between distance and displacement?

Distance is the total length of the path traveled by an object, while displacement is the straight-line distance between an object's initial and final positions. In other words, displacement is the shortest distance between two points, while distance can be longer due to changes in direction.

4. How does acceleration affect velocity?

Acceleration is the rate of change of velocity over time. If the acceleration is positive, the velocity will increase, and if the acceleration is negative, the velocity will decrease. This change in velocity can also affect the direction of motion.

5. Can velocity be negative?

Yes, velocity can be negative if the object is moving in the opposite direction of the chosen positive direction. This indicates that the object is moving backwards or in the opposite direction of its initial position.

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