Question

6. A particle travels 10m during the 4th second and 9.2m during the 6th second .Find the initial

velocity and acceleration?​

Answer 1

Given :

In 1st case :

Distance travelled = 10m

Time taken = 4sec

In 2nd case :

Distance travelled = 9.2m

Time taken = 6sec

To find :

The initial velocity and acceleration

Solution :

Using second equation of kenimatics that is,

s = ut + 1/2at²

where,

• s denotes distance travelled
• u denotes initial velocity
• t denotes time taken
• a denotes acceleration

In first case,

s = 10m and time taken = 4sec.

Substituting all the given values in the equation,

➛ s = ut + 1/2at²

➛ 10 = u(4) + 1/2(a)(4)²

➛ 10 = 4u + 1/2(a)(16)

➛ 10 = 4u + 8a

➛ 2.5 = u + 2a

➛ u = 2.5 - 2a ⠀⠀⠀⠀⠀⠀⠀⠀...(1)

Similarly,

In second case,

s = 9.2m and time taken = 6sec.

Substituting all the given values in the equation,

➛ s = ut + 1/2at²

➛ 9.2 = u(6) + 1/2(a)(6)²

➛ 9.2 = 6u + 1/2(a)(36)

➛ 9.2 = 6u + 18a ⠀⠀⠀⠀⠀⠀⠀...(2)

Now, substituting eq (1) in eq (2),

➛ 9.2 = 6(2.5 - 2a) + 18a

➛ 9.2 = 15 - 12a + 18a

➛ 9.2 = 15 + 6a

➛ 9.2 - 15 = 6a

➛ -5.8 = 6a

➛ a = -0.9 m/s²

Now, put a = -0.9m/s² in eq (1),

➛ u = 5 - 2(-0.9)

➛ u = 5 + 1.8

➛ u = 6.8 m/s

Thus, Initial velocity of the particle is 6.8 m/s and accerlation is -0.9m/s².

Remember !

If the speed is decreasing with time then acceleration is negative. The negative accerlation is called deceleration or retardation.

$\:$

Answer 2

${\huge{ \bf{{\pink{\underline{\underline{\red{\mathbb{\bigstar{\pink{heya \: mate}}}}}}}}}}}$

Given :

In 1st case :

Distance travelled = 10m

Time taken = 4sec

In 2nd case :

Distance travelled = 9.2m

Time taken = 6sec

To find :

The initial velocity and acceleration

Solution :

Using second equation of kenimatics that is,

s = ut + 1/2at²

where,

s denotes distance travelled

u denotes initial velocity

t denotes time taken

a denotes acceleration

In first case,

s = 10m and time taken = 4sec.

Substituting all the given values in the equation,

➛ s = ut + 1/2at²

➛ 10 = u(4) + 1/2(a)(4)²

➛ 10 = 4u + 1/2(a)(16)

➛ 10 = 4u + 8a

➛ 2.5 = u + 2a

➛ u = 2.5 - 2a ⠀⠀⠀⠀⠀⠀⠀⠀...(1)

Similarly,

In second case,

s = 9.2m and time taken = 6sec.

Substituting all the given values in the equation,

➛ s = ut + 1/2at²

➛ 9.2 = u(6) + 1/2(a)(6)²

➛ 9.2 = 6u + 1/2(a)(36)

➛ 9.2 = 6u + 18a ⠀⠀⠀⠀⠀⠀⠀...(2)

Now, substituting eq (1) in eq (2),

➛ 9.2 = 6(2.5 - 2a) + 18a

➛ 9.2 = 15 - 12a + 18a

➛ 9.2 = 15 + 6a

➛ 9.2 - 15 = 6a

➛ -5.8 = 6a

➛ a = -0.9 m/s²

Now, put a = -0.9m/s² in eq (1),

➛ u = 5 - 2(-0.9)

➛ u = 5 + 1.8

➛ u = 6.8 m/s

Thus, Initial velocity of the particle is 6.8 m/s and accerlation is -0.9m/s².

Remember !

If the speed is decreasing with time then acceleration is negative. The negative accerlation is called deceleration or retardation.

$\:$