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Physics Gurus Pls Do Dis by Nobody: 8:33pm On Jul 07, 2015
A car 4m long came up behind a semi-trailer
20m long travelling at a steady rate of 72km/
h. The car driver then overtook the truck. If
the car pulled out from behind the truck with
the front of the car 10m behind the rear of the
truck and accelerated at 3.5 m/s 2 until the
rear of the car was 10m in front of the truck,
how far would the car travel and how long
would it take?
Re: Physics Gurus Pls Do Dis by flames007(m): 8:40pm On Jul 07, 2015
hmm.....dis calls dy/dx + newtons laws.i'll be back!
Re: Physics Gurus Pls Do Dis by marshalcarter: 8:47pm On Jul 07, 2015
Ur question no correct oo
Re: Physics Gurus Pls Do Dis by Pinocchioo(f): 9:16pm On Jul 07, 2015
Queed come and see o
Re: Physics Gurus Pls Do Dis by Nobody: 9:49pm On Jul 07, 2015
marshalcarter:
Ur question no correct oo
wake up from Ur slumber... its correct
Re: Physics Gurus Pls Do Dis by Nobody: 10:53pm On Jul 07, 2015
physics gurus rush in here.. don't be scared.
Re: Physics Gurus Pls Do Dis by olusolaj(m): 10:58pm On Jul 07, 2015
the car would travel by 192.2m with the time of 7.4seconds


Truck speed, St = 72km/h = 20m/s
Distance covered by truck = P
Truck Speed, St =P/t, where t is the time taken
20=P/t,
t=P/20…………(1)
also, Distance covered by car = 10+20+10+4+P
Car Speed, SC =(10+20+10+4+P)/t
Car acceleration, Ac ={(10+20+10+4+P)/t}/t
= (10+20+10+4+P)/t^2
i.e 3.5 = (10+20+10+4+P)/t^2
i.e t^2 = (10+20+10+4+P)/3.5 ………..(2)

sub. The value of t in equation 1 in equation 2
we have, (P^2)/(20^2) = (10+20+10+4+P)/3.5
i.e (P^2)/(400) = (44+P)/3.5
3.5P^2 - 400P - 17600 = 0
Therefore P = 148.21m or -33.93m
Since the car & truck are moving forward, P cannot be negative
Therefore P = 148.21m
Then the distance covered by the car would be = 10+20+10+4+148.21
=192.21m

Also, since the time taken for the car and truck are the same;
time taken , t = 148.21/20 (i.e time, t = distance, P / Speed, St)
= 7.41seconds
Re: Physics Gurus Pls Do Dis by Queed: 10:58pm On Jul 07, 2015
Pinocchioo:
Queed come and see o


who told this one that I am a physics guru?? grin abeg no fall my hand oo grin



Faba:
A car 4m long came up behind a semi-trailer
20m long travelling at a steady rate of 72km/
h. The car driver then overtook the truck. If
the car pulled out from behind the truck with
the front of the car 10m behind the rear of the
truck and accelerated at 3.5 m/s 2 until the
rear of the car was 10m in front of the truck,
[b]how far would [/b]the car travel and how long
would it take?

This your question isn't properly asked angry

though I think I somewhat understand it smiley

@the bolded

o boy, it would travel an infinite distance as long as fuel no finish cheesy cheesy

Had it been you asked
how far HAS the car travelled

hmmm... let's see

10m behind + 20m length of the trailer +

10m in front of the trailer + 4m length of the car + Xm distance travel by the

trailer in time 't' the car took to overtake the trailer (yes, the trailer moved a certain distance too during the time the car used to overtake it)

*Re-modified grin * made unnecessary mistakes grin

to find the distance X
V = X/t

t = -2u+sqrt(1600+8*3.5*44) /2*3.5

t = 1.9sec

72km/h = 20m/s
20= X/1.9

X= 38m

total distance = 44+38= 82m

total time = -2u+sqrt(1600+2296)/2*3.5
T = 3.2sec

here you go...

82m,,, 3.2sec

all calculations were done assuming there was NO relative motion between them AT FIRST smiley

Take this with a pinch of salt tho grincheesy

cc RobinHez ... Teempakguy Oya come and share your thoughts ..
grin grin
Re: Physics Gurus Pls Do Dis by Nobody: 11:23pm On Jul 07, 2015
Faba:
A car 4m long came up behind a semi-trailer
20m long travelling at a steady rate of 72km/
h. The car driver then overtook the truck. If
the car pulled out from behind the truck with
the front of the car 10m behind the rear of the
truck and accelerated at 3.5 m/s 2 until the
rear of the car was 10m in front of the truck,
how far would the car travel and how long
would it take?

olusolaj has gotten the right answer. Pains me that he beat me to it, but, cheesy
What can I say? cheesy
Re: Physics Gurus Pls Do Dis by Nobody: 11:31pm On Jul 07, 2015
olusolaj:
the car would travel by 192.2m with the time of 7.4seconds
How did u arrive at your answer..
Re: Physics Gurus Pls Do Dis by Nobody: 11:34pm On Jul 07, 2015
cc lasticlala fynestboi Richie abeg move dis to FP... I need more answers to be sure of the perfect one plssssssss...
Re: Physics Gurus Pls Do Dis by marshalcarter: 11:39pm On Jul 07, 2015
Faba:
wake up from Ur slumber... its correct
Dats why you dey yab me naw
Re: Physics Gurus Pls Do Dis by marshalcarter: 11:41pm On Jul 07, 2015
Queed:



who told this one that I am a physics guru?? grin abeg no fall my hand oo grin





This your question isn't properly asked angry

though I think I somewhat understand it smiley

@the bolded

o boy, it would travel an infinite distance as long as fuel no finish cheesy cheesy

Had it been you asked
how far HAS the car travelled

hmmm... let's see

10m behind + 20m length of the trailer +

10m in front of the trailer + Xm distance travel by the

bus in time 't' the car took to overtake the trailer (yes, the trailer moved a certain distance too during the time the car used to overtake it)

*modified* made unnecessary mistakes grin

to find the distance X
V = X/t

t = -2u+sqrt(1600+8*3.5*40) /2*3.5

t = 1.7sec

72km/h = 20m/s
20= X/1.7

X= 34m

total distance = 40+34= 74m

total time = -2u+sqrt(1600+2072)/2*3.5
T = 2.9sec

here you go...

74m,,, 2.9sec

all calculations were done assuming there was NO relative motion between them AT FIRST smiley

Take this with a pinch of salt tho grincheesy

cc RobinHez ... Teempakguy Oya come and share your thoughts ..
grin grin


Plsssss gimme ya brain...e be lyk sey ma own need repairsgrin
Re: Physics Gurus Pls Do Dis by Nobody: 11:47pm On Jul 07, 2015
marshalcarter:
Dats why you dey yab me naw
Am sorry
Re: Physics Gurus Pls Do Dis by Nobody: 11:50pm On Jul 07, 2015
Teempakguy:
noted.

I'm not finished yet, but from my preparation, I found, assuming the semi-trailer was stationary, the car would have traveled a distance of 44 meters to overtake it. Since, from 10m from the rear, through 20m which is the length of the trailer, to 10m in front by which it is now the rear of the 4m car that is in front. Add it up.
10+20+10+4 = 44

However, the lorry itself is NOT stationary. Hence my headache. angry
But I will find the answer. Chill. grin
Seriously u are thinking like me... av done all that.
Re: Physics Gurus Pls Do Dis by Nobody: 11:56pm On Jul 07, 2015
Faba:
Seriously u are thinking like me... av done all that.
nothing bad in that. cheesy
I sometimes don't get the answer to my problems too. But that is what everyone else is for.
Re: Physics Gurus Pls Do Dis by Queed: 12:05am On Jul 08, 2015
marshalcarter:

Plsssss gimme ya brain...e be lyk sey ma own need repairsgrin


too much w33d I guess embarassed , check your signature
Re: Physics Gurus Pls Do Dis by olusolaj(m): 12:36am On Jul 08, 2015
Faba:
How did u arrive at your answer..

Truck speed, St = 72km/h = 20m/s
Distance covered by truck = P
Truck Speed, St =P/t, where t is the time taken
20=P/t,
t=P/20…………(1)
also, Distance covered by car = 10+20+10+4+P
Car Speed, SC =(10+20+10+4+P)/t
Car acceleration, Ac ={(10+20+10+4+P)/t}/t
= (10+20+10+4+P)/t^2
i.e 3.5 = (10+20+10+4+P)/t^2
i.e t^2 = (10+20+10+4+P)/3.5 ………..(2)

sub. The value of t in equation 1 in equation 2
we have, (P^2)/(20^2) = (10+20+10+4+P)/3.5
i.e (P^2)/(400) = (44+P)/3.5
3.5P^2 - 400P - 17600 = 0
Therefore P = 148.21m or -33.93m
Since the car & truck are moving forward, P cannot be negative
Therefore P = 148.21m
Then the distance covered by the car would be = 10+20+10+4+148.21
=192.21m

Also, since the time taken for the car and truck are the same;
time taken , t = 148.21/20 (i.e time, t = distance, P / Speed, St)
= 7.41seconds

1 Like

Re: Physics Gurus Pls Do Dis by Queed: 1:02am On Jul 08, 2015
Teempakguy:
noted.

I'm not finished yet, but from my preparation, I found, assuming the semi-trailer was stationary, the car would have traveled a distance of 44 meters to overtake it. Since, from 10m from the rear, through 20m which is the length of the trailer, to 10m in front by which it is now the rear of the 4m car that is in front. Add it up.
10+20+10+4 = 44

So, how long would it take for a car accelerating at 3. 5 from 20m/s to cover 44?

When this time is found, multiply it by the speed of the trailer to find the distance which the trailer traveled. Then having found this, use it to find the total time taken by the car.

However, there is a problem. Finding time while knowing just distance, acceleration, and initial speed transforms into a quadratic equation.
Look,

Using

S= ut + 1/2 at^2

U= 20m/s
a= 3.5
S= 44

==> 44 = 20t + 1/2 x 3.5t^2
==> 44 = 20t + 1.75t^2
==> 7x^2 + 20t - 176

So, queed . . . come and explain yourself. cheesy cheesy

Meanwhile, olusolaj has gotten the right answer. Pains me that he beat me to it, but, cheesy
What can I say? cheesy



I gave a condition bro; [center]ALL CALCULATIONS WERE DONE ASSUMING THERE WAS NO RELATIVE MOTION BETWEEN THEM AT FIRST[/center]
so you can't say I am wrong tongue

Trust me bro, I know how to escape shits like this cheesy
Re: Physics Gurus Pls Do Dis by Nobody: 1:04am On Jul 08, 2015
Queed:




I gave a condition bro; [center]ALL CALCULATIONS WERE DONE ASSUMING THERE WAS NO RELATIVE MOTION BETWEEN THEM AT FIRST[/center]
so you can't say I am wrong tongue

Trust me bro, I know how to escape shits like this cheesy


I knew it. grin
Well, it has been solved.
Shame on us. . . I guess? cheesy
Re: Physics Gurus Pls Do Dis by Queed: 1:24am On Jul 08, 2015
Teempakguy:
I knew it. grin
Well, it has been solved.
Shame on me. . . I guess? cheesy

better grin,,,, at least I did something that's correct in its own sense cheesy
Re: Physics Gurus Pls Do Dis by Nobody: 1:27am On Jul 08, 2015
Queed:


better grin,,,, at least I did something that's correct in its own sense cheesy
you're right. angry
This is a wake up call for me. I've been monopolizing on mathematics. It's time to diversify. angry


*slaps self in face five times!* angry

I will go over my physics textbook again within the week and solve as many as possible problems. angry



Cheers! angry
Re: Physics Gurus Pls Do Dis by Nobody: 1:28am On Jul 08, 2015
olusolaj:

Truck speed, St = 72km/h = 20m/s Distance covered by truck = P Truck Speed, St =P/t, where t is the time taken 20=P/t, t=P/20…………(1) also, Distance covered by car = 10+20+10+4+P Car Speed, SC =(10+20+10+4+P)/t Car acceleration, Ac ={(10+20+10+4+P)/t}/t = (10+20+10+4+P)/t^2 i.e 3.5 = (10+20+10+4+P)/t^2 i.e t^2 = (10+20+10+4+P)/3.5 ………..(2)
sub. The value of t in equation 1 in equation 2 we have, (P^2)/(20^2) = (10+20+10+4+P)/3.5 i.e (P^2)/(400) = (44+P)/3.5 3.5P^2 - 400P - 17600 = 0 Therefore P = 148.21m or -33.93m Since the car & truck are moving forward, P cannot be negative Therefore P = 148.21m Then the distance covered by the car would be = 10+20+10+4+148.21 =192.21m
Also, since the time taken for the car and truck are the same; time taken , t = 148.21/20 (i.e time, t = distance, P / Speed, St) = 7.41seconds
good.... just wanted yo confirm Ur workings..
Re: Physics Gurus Pls Do Dis by Nobody: 1:30am On Jul 08, 2015
tnx for Ur time guys cc olusolaj teempakguy queed
Re: Physics Gurus Pls Do Dis by Nobody: 1:31am On Jul 08, 2015
Faba:
tnx for Ur time guys
cc olusolaj teempakguy queed
don't bother. embarassed
Thanks though.
Re: Physics Gurus Pls Do Dis by Nobody: 1:36am On Jul 08, 2015
Am going to drop some physics question more here during d day...
Re: Physics Gurus Pls Do Dis by Queed: 1:38am On Jul 08, 2015
Faba:
tnx for Ur time guys cc olusolaj teempakguy queed
anytime bro, my monika is very easy to CC grin wink
Re: Physics Gurus Pls Do Dis by olusolaj(m): 1:43am On Jul 08, 2015
Faba:
tnx for Ur time guys
cc olusolaj teempakguy queed

you are welcome sir
Re: Physics Gurus Pls Do Dis by EduRegard: 1:53am On Jul 08, 2015
marshalcarter:
Ur question no correct oo
what we say when we don't get the answer to a question we personally think we solved correctly enough. cheesy
Re: Physics Gurus Pls Do Dis by marshalcarter: 2:15am On Jul 08, 2015
EduRegard:
what we say when we don't get the answer to a question we personally think we solved correctly enough. cheesy
Lol.....@ least I tried grin







So you dey try talk sey I be olodo naw undecided







Anyway sha...na we wegrin
Re: Physics Gurus Pls Do Dis by marshalcarter: 2:20am On Jul 08, 2015
olusolaj:

Truck speed, St = 72km/h = 20m/s Distance covered by truck = P Truck Speed, St =P/t, where t is the time taken 20=P/t, t=P/20…………(1) also, Distance covered by car = 10+20+10+4+P Car Speed, SC =(10+20+10+4+P)/t Car acceleration, Ac ={(10+20+10+4+P)/t}/t = (10+20+10+4+P)/t^2 i.e 3.5 = (10+20+10+4+P)/t^2 i.e t^2 = (10+20+10+4+P)/3.5 ………..(2)
sub. The value of t in equation 1 in equation 2 we have, (P^2)/(20^2) = (10+20+10+4+P)/3.5 i.e (P^2)/(400) = (44+P)/3.5 3.5P^2 - 400P - 17600 = 0 Therefore P = 148.21m or -33.93m Since the car & truck are moving forward, P cannot be negative Therefore P = 148.21m Then the distance covered by the car would be = 10+20+10+4+148.21 =192.21m
Also, since the time taken for the car and truck are the same; time taken , t = 148.21/20 (i.e time, t = distance, P / Speed, St) = 7.41seconds
I bow before you albert einstein

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