Engineering Mechanics - Kinematics of Particle (KOP) - Discussion

Discussion :: Kinematics of Particle (KOP) - General Questions (Q.No.1)

1. 

A race car starting from rest moves along a straight track with an acceleration as shown in the graph (where for t 10 s, a = 8 m/s2). Determine the time t for the car to reach a speed of 50 m/s.

[A]. t = 11.25 s
[B]. t = 6.25 s
[C]. t = 12.5 s
[D]. t = 3.53 s

Answer: Option A

Explanation:

No answer description available for this question.

Veni said: (Dec 21, 2010)  
How answer will come? Please explain.

Abhishek Chouksey said: (Apr 15, 2011)  
Its Simple>
in first 10 sec it reaches only 40 m/s=1/2*10*8(uniform acc)
in next 1.25 sec at constant acc.(v=u+at) it tends to 50 m/s.

Puru said: (Jun 3, 2011)  
Dear, please tell me how it comes 40m/s. By which law and logic you calculate 40m/s.

FYI
1. V=u+at
3. v2=u2+2as
2. s=ut+1/2at2
4. s=1/2(v+u)t

Gopala Krishna said: (Jun 30, 2011)  
Here the car is not moving with uniform accleration upto ten seconds
area of the triangle=0.5*8*10=40m/sec(t=10)
now accleration is constant

So apply v=u+at
50=40+8*t
t=1.25

So total timet = 10+1.25 = 11.25

Mani said: (Aug 25, 2011)  
S=1/2 (8+0) 10.

Since initial velocity is zero. My guess.

Santhosha said: (Aug 31, 2011)  
Initially the is at rest, we consider u=40m/s, and v=50m/s then

v=u+at
50=40+8*t
t=1.25sec,

This is the time required to travel from 40m/s velocity to 50m/s velocity. then the time required to travel the car from rest position is 10+1.25=11.25 sec.

Ankit Bhardwaj said: (Sep 24, 2011)  
That's very simple...

We know dat a= dv/dt
or we can say that v= integral(a dt)...............(1)
Now by graph we can see that a=(4/5)t upto the first 10 seconds
Now put v=50 in the eq.(1)

So after calculating t= 11.25 sec.

Sekhar said: (Sep 15, 2012)  
Actually we want calculate and differentiating function, like ankith what he was explain.

Wasprulez1990 said: (Oct 7, 2012)  
The area in the graph is velocity i.e. v=at, so the total area is given i.e.final velocity 50 m/s,

So 50= area of triangle + area of rectangle.

50= (0.5 x base x height) + (width x height).

50= (0.5 x 10 x 8) + [ (t-10) x 8].

Finally t=11.25 seconds.

Aishwarya Katale said: (Dec 2, 2013)  
How to calculate 40? which is v?

Ankit Kumar said: (Oct 27, 2014)  
Aishwarya Katale : Actually 40m/s its area under the curve. :-).

Manan said: (Dec 7, 2014)  
The answer is incorrect correct answer is 6.25s.

Solution u = 0.

v = 50 m/s.
a = 8 m/s^2.

So just find t by v = u+at.
Which is 50 = 0+8t?

50 = 8t.
t = 50\8 = 6.25s.

Hasya said: (May 17, 2016)  
No, 11.25 is correct.

The formula v = u + at is applicable only in the case of constant acceleration.

Since up to the first 10 seconds the car is moving wth increasing acceleration area under the triangle gives final velocity at t =10 sec since initial velocity is 0.

Kumara Swamy said: (May 31, 2016)  
11.25 secs are the correct answer.

Equations of motion are:

v = u + at;.

s = ut + 0. 5 at^2 etc are valid when accleration is constant.

In the question, the acceleration varies with timeup to 10 sec and then it is constant so, use
v = u + at after 10 secs.

And for before 10 secs use an area under a - t graph is the velocity similarly area under v - t graph gives displacement.

You will understand this concept when you pass through basics of calculus.

Srikanth said: (Aug 10, 2016)  
Acceleration * time = velocity.
So,in the above graph find the area covered by line up to uniform acceleration.

i.e., Area of triangle = 1/2 * 10 * 8 = 40 m/s. And then it follows constant acceleration and the required area(rectangle) is 10. So 8 * 1.25 = 10 m/s, so a total of (40 + 10)m/s is achieved in (10 + 1.25)secs.

Ss Ahamed Halith said: (Oct 10, 2019)  
1. We consider the given acceleration and time (i.e a=8 ms-2 t=10s) then we find velocity for that next we pick any one choices given and add or sub for making it equal to 10 find another velocity.

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