Question

A body travels 180cm in the first three seconds and 220cm in the next 5seconds. What will be the velocity at the end of the ninth second?

Answer 1

A body travels 180 cm in the first three seconds and 220 cm in the next 5 seconds.

s= ut + 1/2 at²

In first there seconds, s= 180 cm = 1.8 m and t = 3 sec

1.8 = u(3) + 1/2 a(3)²

1.8 = 3u + 1/2 × 9a

1.8 = 3u + 9a/2

3.6 = 6u + 9a

1.2 = 2u + 3a

1.2 - 3a = 2u...............(1)

Now, in next 5 seconds, s= 220+180 = 400 cm = 4 m and t = 5+3 = 8 sec

4 = u(8) + 1/2 a(8)²

4 = 8u + 64a/2

4 = 8u + 32a

1 = 2u + 8a

1 - 8a = 2u….................(2)

On comparing eq (1) & (2) we get,

1 - 8a = 1.2 - 3a

1 - 1.2 = -3a + 8a

-0.2 = 5a

a = -0.04 m/s² [ a = Acceleration ]

Substitute value of a in eq (2)

1 - 8(-0.04) = 2u

1 + 0.32 = 2u

u = 0.66 m/s [ u = Initial velocity ]

We have to find the velocity at the end of the ninth second.

Now, v = u + at and here, t = 9 sec

v = 0.66 + (-0.04)(9)

v = 0.66 - 0.36

v = 0.3 m/s [ v = Final velocity ]

Answer 2

Given:-

A body travels 180 cm in the first 3 seconds and 220 cm in the next 5 seconds.

To find:-

The velocity of the body at the end of the ninth second.

Solution:-

180 cm = 180/100 = 1.8 m

220 cm = 220/100 = 2.2 m

In first 3 seconds,

s = 1.8 m

t = 3 s

Using the formula:

$\displaystyle{s = ut+\frac{1}{2}at^2 }$

1.8 = u*3 + 1/2 a(3)²

⇒1.8 = 3u + 1/2 9a

⇒3u = 1.8 - 1/2 9a

⇒6u = 3.6 - 9a

⇒2u = 1.2 - 3a                            ...(i)

In next 5 seconds,

s = 2.2 m + 1.8 m = 4 m

t = 5 s + 3 s = 8 s

Using the formula:

$\displaystyle{s = ut+\frac{1}{2}at^2 }$

4 = u(8) + 1/2 a(8)²

⇒4 = 8u + 1/2 64a

⇒8u = 4 - 1/2 64a

⇒2u = 1 - 8a                        ...(ii)

On comparing eq.(i) and eq.(ii), we get:

1 - 8a = 1.2 - 3a = 2u

⇒1.2 - 1 - 3a + 8a = 0

⇒0.2 + 5a = 0

⇒a = -0.2/5

⇒a = -0.04

Substituting the value of 'a' in eq.(ii) :

2u = 1 - 8a

⇒2u = 1 - 8(-0.04)

⇒2u = 1 - (-0.32)

⇒2u = 0.32 + 1

⇒2u = 1.32

⇒u = 1.32/2

⇒u = 0.66 m/s

Now, t = 9 sec

We know,

$\displaystyle{v=u+at}$

⇒v = 0.66 + (-0.04)9

⇒v = 0.66 - 0.36

⇒v = -0.3 m/s

∴So, the velocity at the end of the ninth second is 0.3 m/s or 30 cm/s.