Question

Derived 3=Ut+^12at^2 using graphical method. where S,U,t & a usual meaning.
its really hard for me. pls help me. ​

Answer 1

Explanation:

Consider the velocity-time graph of a body shown in the figure. The body has an initial velocity u at a point A and then its velocity changes at a uniform rate from A to B in time t. In other words, there is a uniform acceleration a from A to B, and after time t its final velocity becomes v which is equal to BC in the graph. The time t is represented by OC.

Suppose the body travels a distance s in time t. In the figure, the distance traveled by the body is given by the area of the space between the velocity-time graph AB and the time axis OC, which is equal to the area of the figure OABC.

Thus:

Distance traveled = Area of figure OABC

= Area of rectangle OADC + area of triangle ABD

Now, we will find out the area of rectangle OADC and area of triangle ABD.

(i) Area of rectangle OADC=OA×OC

=u×t

=ut

(ii) Area of triangle ABD=

2

1

×Area of rectangle AEBD

=

2

1

×AD×BD

=

2

1

×t×at

=

2

1

at

2

Distance travelled, s= Area of rectangle OADC + area of triangle ABD

s=ut+

2

1

at

2

Answer 2

Answer:

(a) Suppose a body has an initial velocity 'u' and a uniform acceleration 'a' for time 't' so that its final velocity becomes 'v'. Let the distance travelled by the body in this time be 's'. The distance travelled by a moving body in time 't' can be found out by considering its average velocity. Since the initial velocity of the body is 'u' and its final velocity is 'v', the average velocity is given by 

Average velocity =2Initial velocity + Final velocity

That is, Average velocity =2u+v 

Also, Distance travelled = Average velocity × Time 

So, s=(2u+v)×t

From the first equation of motion, we have, v=u+at.

Put this value of v in equation (1), we get: 

s=(2u+u+at)×t

or s=2(2u+at)×t

or  s=22ut+at2

or s=ut+21at2

where, s= distance travelled by the body in time t

u= initial velocity of the body 

and a= acceleration 

(b) Initial velocity, u=0m/s

Final velocity, v=36km/h=10m/s

Time, t=10min=10×60=600 sec

Acceleration =time takenFinal velocity - Initial velocity

So, a=tv−u=