A car starts from rest and accelerates uniform upto speed 90 km/hrs in 5 sec. Total distance covered

RAKESH   9 years AGO

v = 90*5/18 =25 m/s

a = v/t = 25/5 = 5 m/s^2

d = 1/2 * at^2 = 62.5 m

Like? Yes (13) | No (1)  

BHASKAR MANDAL   9 years AGO

v = u+at
a = 5 m/s^2
s = ut + 1/2 a*t^2
s = 62.5

Like? Yes (3) | No   

Saurabh Kumar Gupta   9 years AGO

90kmph = 25mps
so a = (v-u)/t = (25-0)/5 = 5
s = ut + (at^2)/2 = 0*5 + (5*5*5)/2 = 62.5

Like? Yes (3) | No (1)  

Supriya Mondal   9 years AGO

Distance=ut+(0.5)ft^2
u=0,t=5,f=5
After putting the values we get=62.5
So,option 6)62.5 is the right answer

Like? Yes (3) | No (1)  

ANKT   9 years AGO

given u=0 t=5sec a=? v=90km/hr i.e. 25m/s S=?
now
v=u+at
on calculatin we get a=5m/s^2
now putting values in S=ut+.5at^2
we get S=62.5 m

Like? Yes (2) | No   

Santosh Kumar   9 years AGO

90 km/hr in 5 sec
25 m/sec in 5 sec
a=5m/sec^2(by formula v=u+at)
s=1/2a*t^2
s+1/2*5*5*5=62.5

Like? Yes (2) | No (1)  

Abhishek birkh   9 years AGO

accelaration=(v-u)/t=5m/s
s=ut+1/2 at*t=62.5m

Like? Yes (1) | No   

apurva katyayni   9 years AGO

u= 0, v=90km/h=25m/s
a=(v-u)/t=25/5=5
s=ut + (1/2) at^2= 125/2 = 62.5m

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ABHINAV PUSHPAM   9 years AGO

Acceleration=(v-u)/t
So, 90/5= 5 m/sec
Distance= 125/2= 62.5

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MRITUNJAY   9 years AGO

s=u*t+1/2a*t^2
a=v-u/t,v=90*5/18=25m/s
a=25-0/5=5m/s^2
u=intial velocity,v=final velocity
u=0,t=5
s=0*5+1/2*5*25
s=125/2
s=62.5
option 6.

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manohar kumar singh   9 years AGO

using formula
v=u+at
s=ut+ 1/2*a*t*t
putting u=0.
s is distance=62.5.

Like? Yes (1) | No   

sravani   9 years AGO

v=u+at
v=90kmph=25mps
u=0
t=5
so,a=5mps^2
s=ut+1/2 at^2
s=62.5m

Like? Yes (1) | No   

Surit Datta   9 years AGO

according to the formula S=ut+1/2*f*t*t
we get the appropriate result.

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sompon temtheerakij   9 years AGO

speed 90 km/hr = 25 m/sec
distance = area under curve = 25*5/2 = 62.5 m.

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Ajinkya Kulkarni   9 years AGO

125/2=62.5

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Rahul Das   9 years AGO

v=u+ft
u=0
f=5
s=1/2*f*t*t
=1/2*5*25
=62.5

Like? Yes (1) | No   

SURABHI JAISWAL   9 years AGO

u=0, t=5 sec,v=90 km/hr
a=(v-u)/t
a=1/200(km/s^2)
s=ut +1/2(at^2)
s=62.5m

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Aditya Raghuwanshi   9 years AGO

v=u+at^2
so..at initial u=0
a=90km/hr
t=5 sec

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Swathy Murugan   9 years AGO

the acceleration is 5
so the distance = 1/2(a(t^2))=1/2*5*25=62.5

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Satya B   9 years AGO

u=0 v=90 t= 5
a= (v - u)/t
= 90/5
= 18kmph= 5 mpsec
d= u+ 1/2 a t^2
= 0+1/2 * 5 *25 = 62.5 m

Like? Yes (1) | No   

modem mounika   9 years AGO

a=(v2-v1)/(t2-t1) = (25-0)/(5-0)=5m/s
v2=25,v1=0
v2^2-v1^2=2a(s2^-s1) (I.e, s1=0)
s2=62.5

Like? Yes (1) | No   

POOJA SINGH   9 years AGO

u=0 then from v= u+ at we get a = 5 m^2/s
s=ut +1/2 at^2 so we get s= 62.5

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Anshul Garg   9 years AGO

Using v = u + at
we can find the value of a i.e. acceleration
where u = 0 and t =5 and v = 25 (Given in the question)

Now using s = ut + 1/2at^2 we can get the distance and answer is 62.5

Like? Yes (1) | No   

Testing Account   9 years AGO

avg speed--0 90/2=45 then convert into m/second by multiply 5/18 then we get 62.5m/s

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Giridharan Manokaran   9 years AGO

In an acceleration mode, the average velocity will be average of initial and final velocity.

So, in this question
initial velocity=0
final velocity=90kmph
average velocity is (0+90)/2=45kmph.

At this velocity the distance covered by car for 5 seconds will be,
distance=velocity*time=45kmph*5=(45*1000/3600)*5=62.5

So, the distance covered by the car is 62.5 meter

Like? Yes (1) | No   

Dharmraj Sharma   9 years AGO

acceleration= (final velocity- initial velocity)/time
a=(25-0)/5=5 m/(s*s)
s=ut+1/2 a*t*t
s=0*5+1/2 5*5*5=62.5m

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Rahul Hooda   9 years AGO

Converting speed in m/s 90km/h = 25m/s
Acceleration = (25-0)/5 = 5m/s^2
By 3rd Equation of motion S = ut + (at^2)/2
S = 0x5 + (5x5x5)/2
S = 62.5

Like? Yes (1) | No   

neethu   9 years AGO

u=0,v=90km/hr=25m/s
a=((v-u)/t)
a=((25-0)/5)=5m/s2
d=(ut+((1/2)at^2)))
substitute all values
d=62.5m

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Vivek Vishal Anand   9 years AGO

25=0+5a
a=5
S=0+(1/2)5*25
S=62.5

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kumari neha   9 years AGO

We know that v=u+at;
where v=final velocity,u=Initial Velocity;
Given u=0 m/sec, v=90 Kmph=25 m/sec and t=5 sec
and Distance(S)=ut+0.5at^2
So from eq 1...a=(25-0)/5=5 m/sec^2
So S=0*t+0.5*5*5*5=62.5
Answer is S=62.5 m

Like? Yes (1) | No   

Sravanthi   9 years AGO

d=s*t
speed=25m/s
t=5
acceleration a= s/t=5
distance=(at^2)/2
d=62.5

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Ankush Surana   9 years AGO

a=5m/s^2 Then
S=ut+1/2at^2
S=0*5+1/2*5*5^2
S=0+1/2*125
S=62.5 m

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Deepesh joshi   9 years AGO

First equation of motion gives us acceleration a=5

And third equation of motion gives distance

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shubham mishra   9 years AGO

v=u+at
s=ut+1/2at2

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Abhinav Arpanam   9 years AGO

v^2 + u^2 = 2as

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Sandip Biswas   9 years AGO

f=v/t
=(90*5/18)/5
=5 m/sec
s=ut+1/2ft^2
=0+ 1/2*5*5^2
=125/2
=62.5

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harvinder   9 years AGO

v = 90km/h =25m/s

a= (v-u)/t = 25/5 =5

D = (v^2-u^2 ) 2a = 25*25/10 =62.5

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Sourabh Prasad   9 years AGO

90 km/hr = 25 m/s
v=u+at
25=0+a*5
a=5

s=ut+ 1/2 at*t
s =0 + 1/2 5*5*5
s= 125/2
s=62.5 m

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Chandra chur anand   9 years AGO

v=90 km/hr= 25 m/sec
v-u=a*t so here u=0 m/s
so a=v/t=25/5=5 m/s^2
so d= 0.5*a*t^2
=0.5*5*25
=62.5 m is ans....

Like? Yes (1) | No   

Pankaj   9 years AGO

speed in m/s = 90 x (5/18) = 25 m/s
as car starts from rest initial velocity u = 0 m/s
final speed of car = 25 m/s
therefore acc. = (v-u)/t
=(25-0)/5
a=5m/s^2

s = ut + (1/2 a t^2)
= 0 x 5 + (1/2 x 5 x 5 x 5)
= 62.5 m

Like? Yes (1) | No   

Nupur   9 years AGO

The newton's equations of motion can be used here:
Final velocity = 90km/hr = 25m/sec

a = 25/5 = 5m/sec^2
s = 1/2 *a *t^2
Ans = 125/2 = 62.5m

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kamlesh kumar   9 years AGO

s=ut+(at^2)/2,
a={90*1000/(3600)}/5,
t=5,
u=0.

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ravi   9 years AGO

accerlation=25/5=5

s=1/2*5*25=62.5

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Ankit Pandey   9 years AGO

u = 0
v = 90 km/hrs = 25 m/sec

now
v = u + a*t
a = 5 m/ sec^2
now distance
s = ut +(1/2)a*t^2
s = 125/2
s= 62.5 m

Like? Yes (2) | No (1)  

Kiran Kumar reddy   9 years AGO

d=s*t
speed=25m/s
acceleration a=5
distance=(speed)^2/(2a)
d=62.5

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rakhi kumari   9 years AGO

(v-u)/t=a v=90*5/18=25 m/s
a=5 m/s^2
s=v^2/2*a
= 25*25/2*5
= 62.5

Like? Yes (1) | No   

saurav   9 years AGO

u=0
v=90km/hr=25m/s
t=5s

v=u+at
=>a=5m/s^2

s=.5*a*t*t
=>s=62.5m

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anusha   9 years AGO

ans is 62.5

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KUMAR HARSH   9 years AGO

here u=0 and v=90*5/18=25m/s t=5 sec
a=v/t=5m/s^2
s=1/2*a*t^2=125/2=62.5

Like? Yes (1) | No   

Vishakha Tinkhede   9 years AGO

according to v-u = at ans s = ut +1/2at2 , laws of motion
in the question u = 0, v = 90km/hr , t = 5sec
thus a = (v-u)/t which comes out to be 5 . and then using second equation s = 1/2x5x(5)squ. = 62.5

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nikhil   9 years AGO

a=5
s=u*t +0.5*a*t*t

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Corinna   9 years AGO

90/3600=0.025 0.025*5/2=0.0625 km

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Jaspalsingh Parmar   9 years AGO

option 6 is correct

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rakesh yadav   9 years AGO

s-d/t

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Abhishek singh   9 years AGO

its simple if we remember physics formula
v=u+at
now v=90
u=0
t=5 so a=18 km/hr
=>5 m/sec
s=1/2at^2
=>1/2(5*5^2)
=125/2
=62.5

Like? Yes (1) | No   

sumit kumar   9 years AGO

from question u=0 t=5 v=90km/hr=25m/s
acceleration a=(v-u)/t=5m/t^2
total distance covered by uniform a
1/2*a*t^2=62.5

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vishal kumar   9 years AGO

v=u+at which give a=5
V=5*(18/5)=5m/s
so a=5m/sec sq
now
s=ut+1/2at^2 which give s=62.5m ans

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adityakumarsingh   9 years AGO

90*5/18=a*5
a=5
s=0.5*a*5^2
s=125/2
s=62.5

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Paras Jain   9 years AGO

its 62.5 Ans.

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Ankit Sirohi   9 years AGO

v=u+at which give a=5
s=ut+1/2at^2 which give s=62.5m

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JEETENDER KALACHAND KEWLANI   9 years AGO

a=(v-u)/t = (90*1000)/(5*60*60)

a=5 m/sec square

now distance = 0.5*a*t^2
hence distance = 0.5 * 5 * 5^2 = 62.5 m

Like? Yes (1) | No (1)  

kaushik mishra   9 years AGO

As car start from rest so initial velocity u=0,and by equation of motion v=U+at so a=25/5=5m/s.and for the distance vsquare=usquare+2as.so s = vsquare /2a=625/10=62.5

Like? Yes (1) | No (1)