Question

A body describes 400 cm in 3rd second and 1200 cm in the 5 th second. If the motion is uniformly accelerated , how far will it travell in next 3 second ?.

Answer 1

Answer:

Let initial velocity be u and acceleration be a

sn = u + ½a(2n-1)

a/q

u + ½a(2×3-1) = 4  ---1.

u + ½a(2×5-1) = 12  ---2.

Eqn. 2 – eqn. 1

=> ½a(2×5-1) - ½a(2×3-1) = 8

=> 4a = 16

=> a = 4 ms-2

By 1.

u + ½ ×4×5 = 4

=> u = -6 ms-1

Total distance covered 8 s

= -6×8 + ½×4×(8)2

= 80 m

Distance covered in 5th second

= -6×5 + ½×4×(5)2

= 20 m

So distance covered in next 3 second after 5th second will be

80 – 20 = 60 m

Answer 2

Given :

Distance travelled in 3rd second , $s_3=400\ cm=4\ m$ .

Distance travelled in 5rd second , $s_3=1200\ cm=12\ m$ .

To find :

The motion is uniformly accelerated and how far will it travel in next 3 seconds.

Solution :

We know , displacement in nth second is given by :

$s_n=u+\dfrac{a(2n-1)}{2}$

So ,

$s_3=u+\dfrac{a(2\times 3-1)}{2}\\\\4=u+\dfrac{5a}{2}$     ......( 1 )

Also ,

$s_5=u+\dfrac{a(2\times 5-1)}{2}\\\\12=u+\dfrac{9a}{2}$      .......( 2 )

Subtracting eq 2 by 1

We get :

$2a=8\\a=4\ m/s^2$

Also , u = -6 m/s .

Since , the acceleration is constant .

Therefore , the motion is uniformly accelerated .

Now , distance travelled in next three seconds is given by :

$d=s_8-s_5\\\\d=-6+\dfrac{4\times (2\times 8- 1)}{2}-12\\\\d=-6+30-12\ m\\\\d=12\ m$

Therefore , distance travelled in next 3 seconds is 12 m .

Learn More :

Kinematics

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