Kinematics Problems: Truck & Automobile

[redsox5]

I just wanted to double check on my answers here. I have all the problems posted below with my work. If you could make sure I'm doing these right I would greatly appreciate it.

Homework Statement

A truck passes a point A on a straight level road with a constant velocity of 10 m s-1. At the same instant, an automobile starts from rest at point A and travels in the same direction as the truck, with a constant acceleration of 1.6 m s-2.

A) How long a time does the automobile take to pass the truck, from the instant it starts at point A?

B) How far beyond the starting point will the automobile overtake the truck?

C) Determine the average speed of the bus during the interval, t=0 to t=10 s.

D) How fast will the automobile be traveling, when it passes the truck?

E) If the speed limit on this road is 50 km/hr., is the automobile within the speed limit as it passes the truck?


My work is attached below. I did a problem similar to this and found that the answer for my time was half what it should have been.This messed up all my other answers. I'm not sure where I'm making my mistake in finding the time.

 

[nicktacik]

I can't see your attachment yet (pending approval), but the time for question A) should be 12.5s. What did you get?

 

[redsox5]

I got 12.5 once i doubled my answer. I'm not sure why. I used the formula final velocity = oringal velocity + accel. x time

that got me 6.25 (doubled is 12.5 sec)

my other answer were

B) 125m
C) 16 m/s
D) 20 m/s
E) yes

 

[nicktacik]

The equation to find the time should be

$$\frac{1}{2}at^2 = vt$$

Leading to the solution t = {0,2v/a}
You probably forgot the 1/2 in the equation for accelerated motion.

Looks like you are also forgetting that factor of 1/2 for question C.

You might want to try question E again. What is 50km/h in m/s?

 

[redsox5]

For C I should find the distance first whic is 80m in the first10 sec

then use v = d/t

then that gives me 8m/s

That would mean for E

50 km/h = 180 m/s
so that would make the answer No


The answers look good though?

 

[nicktacik]

8 m/s is correct. Everything else looks good except E. 50 km/h is not 180 m/s.

Use dimensional analysis

(50km/h)(X m/km)(h / X min)(min / X s)

Fill in the X's, and multiply through to get your answer in m/s.

 

[redsox5]

13.8 m/s
So that would be yes
I think I'm better at converting now that you broke it down for me.

Do you think you could look at one more problem. I think I might have did a little better on it but I just want to be sure. If you don't have time o problem. Thanks

(I'll just post my scanned work)

 

[nicktacik]

Sure thing.

 

[redsox5]

A ball is thrown vertically upward from the top edge of a building 250 m tall with an initial velocity of 25 m/s. Assume the acceleration due to gravity is g = 10 m s-2.

A) How long will it take for the ball to rise to the highest point on its trajectory?

2.5 sec

B) What is the maximum height, measured from the ground, to which the ball rises?

281.25m

C) On its way up, when is the speed of the ball equal to half its speed of projection ?

1.25 sec

D) Assuming that the ball misses the building on its way down, at what time does the ball hit the ground ?

5 sec

E) What is the magnitude of the velocity with which the ball hits the ground?

55 m/s (I got this answer from D, I couldn't fit it on the page. This comes from V2)

 

[HallsofIvy]

You have, for "A)How long a time does the automobile take to pass the truck, from the instant it starts at point A?",
$v_f= v_0+ at$
10 m/s= 1.5 m/s² t
t= 6.25
and then say "I used 12.5 because that's 2 times 6.25"! Without giving any reason for doubling! Even if it is true that you got 1/2 the correct value before, that is no reason for arbitrarily doubling whatever you get!

In any case, you are answering the wrong question! v_f= v₀+ at says two speeds are the same. The answer you got is the time after which the two vehicles have the same speed, not the same distance, which is what is asked. Since the car starts out going slower than the truck, at the time the speeds are finally the same, the car will be well behind the truck still.

Yes, you could argue, since this is a constant speed versus a constant acceleration, the distance the car loses while it is slower than the truck will be offset by the distance the car gains while it is faster than the truck- by "symmetry" the total time should be twice the time until they have the same speed- but at least you know now why you multiply by 2 (and when that will not work).

A better, clearer way to do this is to use distance formulas rather than velocity formulas. The truck, moving at constant v_t will go v_t t meters in t seconds. The car, at constant acceleration, will go a distance of (1/2)at² meters in t seconds. The car and truck will have gone the same distance when v_t t= (1/2)at². Notice that that has the obvious solution t= 0. If t is not 0, we can divide both sides by t to get v_t= (1/2)at. Now put in the numbers you are given and you will get the correct answer- but do you see why you had to double your first answer?

"B) How far beyond the starting point will the automobile overtake the truck?"
Once you have the answer from (A), put that value of t into either side of the equation for distance. I notice you used (1/2)at². You would get the same answer, and the arithmetic is easier, using v_t t.

"C) Determine the average speed of the bus during the interval, t=0 to t=10 s."
I have no idea how to answer this since there is no bus in the problem! Assuming you meant "automobile", divide the distance it went that you got in (B) by the time you got in (C). For some reason, you use 10 seconds as the time, even though you got 25 seconds as the time in (A). What you have calculated is the speed of the automobile after 10 seconds. I have no idea why you think that would be the average speed.

"D) How fast will the automobile be traveling, when it passes the truck? "
Put the time you got in (A) into the formula for speed with constant acceleration. What you have is correct.

"E) If the speed limit on this road is 50 km/hr., is the automobile within the speed limit as it passes the truck?"
You multiplied 20 m/s by 3.6 and got 72 km/hr and answered "Yes".
First you should say WHERE you got that "3.6". It is (almost) correct but you should show where you got it.
The reason I said "almost" is that "20 m/s times 3.6" is NOT "72 km/hr"- it is 72 m/s. 20 m/s times 3.6 (km/m)(second/hr) is 72 km/hr.

Finally, if you think going 72 km/hr is "within the speed limit" of 50 km/hr then I don't want to be on the same road with you!

 

[nicktacik]

Everything looks good except for E.

The equation for V(t) is V(t) = 25m/s - (10m/s)t
So evaluate the magnitude of V(5)

 

[redsox5]

So the magnitude for E would be 3?

 

[nicktacik]

Sorry, it looks like your answer for D is also wrong. Which of course screws up the answer for E. I will wait until I can see your work.

 

[redsox5]

Thanks for the explanation halls. I just needed someone to explain exactly that for me. Now I'm coming up with 15 s for A.

I did:

12 m/s (t) = .5 (1.6 m/s) t^2
t = 15 sec

 

[nicktacik]

Why 12m/s?

"A truck passes a point A on a straight level road with a constant velocity of 10 m s-1"

 

[redsox5]

ok, I got it mixed up with another problem

 

[redsox5]

for problem 2 here's my work

final V = original V + a x delta T

V2= 30 m/s + 10 m/s^2 x 2.5s
V2 = 55 m/s

then i used the formula:

t = V2-V1 / a

t = (55 m/s - 30 m/s) / 10 m/s^2
t = 2.5 sec

I multiplied that by 2 since if took 2.5 to reach the top

 

[redsox5]

After running through problem 1 again I got:

A) 12.5 sec
B) 125 m
C) 12.5 m/s
D) 20 m/s
E) NO

Is that more like it. Thanks for the help.

 

[nicktacik]

Ok there's your problem. You are using a to be +10m/s^2. However, you throw the objects up, and gravity pulls it down, so you want a negative sign. A more suitable equation of motion would be

y(t) = 250 + 25t - (1/2)(10)t^2
v(t) = 25 - 10t

Using this for D, you can use y(t) = 0 (when it hits the ground), then find the time it takes. For E, substitute that value into the v(t) equation.

 

[redsox5]

I think I found my mistake.

t = (281.25 m/s) / (10m/s^2)
t = 28.125 sec

28.125 sec + 2.5 sec = 30.625 sec

That would make E:

V2 = V1 + a(delta t)
v2= 10 m/s^2 (28.125 sec)
v2 = 281.25 m/s

 

[redsox5]

I'm coming up with 45 sec now, it may seem a little high

 

[nicktacik]

Keep at it, 10s should be the correct answer. (for D)

 

[redsox5]

so the magnitude of the velocity is negative

 

[nicktacik]

The velocity is negative since common sense tells us the ball will be going down when it hits the ground after it is thrown from a higher height. However, when they say the magnitude of the velocity, they mean the absolute value, so the magnitude will be positive.

 

[redsox5]

thanks, you think everything else looks good?

 

[nicktacik]

The wording for problem 1c. makes me think the answer should be 8s.
#2 a-c look good.

Do you understand how I got 10s for 2d? And then, what should that give for 2e?

 

[redsox5]

yes, and i got 75 for 2e, let me look over 1c again

 

[redsox5]

how are you coming up with 8 s?

 

[nicktacik]

Okay good, 75m/s is correct.

Woops meant 8m/s

"C) Determine the average speed of the bus(automobile) during the interval, t=0 to t=10 s."

Not sure, why 10 is chosen, but we'll go with it. X = 1/2at^2 = (1/2) * (1.6m/s^2) * (10s) ^ 2 = 80m

Average velocity = 80m/10s = 8m/s