Calculating Distance and Velocity in a Moving Object Scenario

[nblu]

Regarding Forces & Motion (edited)

Hi, thank you for your time to read this thread.
I'll begin with the question, thanks again!

Q: At the instant a traffic light turns green, a car starts from the rest and accelerates
uniformly at a rate of 4.0 m/s^{2} East. At the same instant, a truck traveling with a
constant velocity of 90.0 km/h East overtakes and passes the car.

a) How far beyond the starting point is the car after 10.0 s?
b) How far beyond the starting point is the truck after 10.0 s?
c) The car passes the truck at a distance of 312.5m beyond the starting point.
How fast is the car traveling at this instant?
d) How long does the car take to pass the truck?
e) Both drivers suddenly see a barrier 100.0m away and hit their brakes at exactly
same time. Assuming that both vehicles decelerate uniformly and they take 3.0 s
to stop, will the stop in time?


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Here are my answers :)
A: (I've omit vector signs, not sure how to input the symbol)
a) a = 4.0 m/s^{2}, \Deltat = 10.0 s
\Deltad = (v)(t) + 1/2(a)(t^2)
\Deltad = (0)(10) + 1/2(4)(100)
\Deltad = 200m

b) v = 90.0 km/h (convert to m/s which equals to 25m/s), \Deltat = 10.0 s
\Deltad = (v)(t)
\Deltad = (25)(10.0)
\Deltad = 250m

c) \Deltad = 312.5m, a = 4.0 m/s^{2}
vii^{2} = vi^{2} + 2a\Deltad
vii = (0) + 2(4.0m/s^{2})(312.5m/s)
vii = 50 m/s

d) a = 4.0 m/s^{2}, vii = 50 m/s
\Deltat = (vii - vi) / a
\Deltat = (50m/s - 0) / 4.0 = 12.5 seconds

e) Here, what I thought of doing was, find both car and truck's acceleration
then find their distance within 3 seconds then check whether the answer is
higher than 100m or not.

***Car
a = \Deltav / t
a = 0 m/s - 50 m/s / 3.0s
a = -16.7 m/s^{2}

\Deltad = (vii)\Deltat - 1/2a\Deltat^{2}
\Deltad = 75.15m

***Truck
Same calculation
\Deltad = 37.35m

// So I've said that those two vehicles will stop on time because the calculated
distance is less than 100m.

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Thank you for reading this.. I really appreciate it.. It's very long..
I needed some advice especially on e), because that question really got me thinking.
I honestly don't have confidence that my answers are right :( If I'm wrong
please correct me XD

Thanks again!

 

[malty]

Hi, thank you for your time to read this thread.
I'll begin with the question, thanks again!

Q: At the instant a traffic light turns green, a car starts from the rest and accelerates
uniformly at a rate of 4.0 m/s^2 East. At the same instant, a truck traveling with a
constant velocity of 90.0 km/h East overtakes and passes the car.

a) How far beyond the starting point is the car after 10.0 s?
b) How far beyond the starting point is the truck after 10.0 s?

*******************************

For question a), is it be possible to use a = v/t equation to find the velocity
then use v=d/t to find the final answer?

I was wondering if the same thing happens for question b) it looks much simpler
if use v=d/t equation.

Any advice would be appreciated, many thanks!

a) It may be using complex math, but otherwise no.
how about using the formula s=v_i t+\frac{at^2}{2}

b)You know it's velocity, which is constant, so how would you calculate distance over a certain time interval?

Remember to use SI units.

 

[nblu]

a) It may be using complex math, but otherwise no.
how about using the formula s=v_i t+\frac{at^2}{2}

b)You know it's velocity, which is constant, so how would you calculate distance over a certain time interval?

Remember to use SI units.

So the V in that equation represents the initial, which is 0 right? t would be the given unit
in the question and same for b)

Thanks!