Question

Answer with steps
v=A+Bt+Ct^2
v-velocity
t-time
units of A,B and C?

F=$\frac{a}{t}$+bt^2
F-force
t-time
dimensions of a and b?

v=at+$\frac{b}{(t+c)}$
v-velocity
t-time
dimensions of a,b and c?

Answer 1

Solution :

(i) : Given relation,

v = A + Bt + Ct²

As LHS represents velocity , Each term of RHS must represent velocity.

[ ∵ Velocity can be added only to Velocity ]

Therefore ,

⇒ A = v = m/s

And,

Bt = v or B = v/t

Writing the units of v and t , We get

B = (m/s) / s

⇒ B = m/s²

And,

Ct² = v or C = v/t²

Writing the units of v and t , We get

C = (m/s) / s²

⇒ C = m/s³

Hence,

A's unit = m/s , B's unit = m/s²

C's unit = m/s³

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(ii) : Given relation,

F = a/t + bt²

As LHS represents Force , Each term of RHS represent Force.

[ ∵ Force can be added only to Force ]

Therefore ,

a/t = F or a = F × t

Writing the Dimensions of F and t , We get

$\text{a} = [ ML{T}^{ - 2} ] \times [T] \\ \\ \boxed{ \: \text{a} = [ ML{T}^{ - 1} ] \: }$

And,

bt² = F or b = F/t²

Writing the dimensions of F and t, We get

$\text{b} = \frac{[ ML{T}^{ - 2} ]}{ [{T }^{2} ]} \\ \\ \boxed{ \: \text{b} = {[ ML{T}^{ - 4} ]} \: }$

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(iii) : Given relation,

v = at + b / (t + c)

As time can be added only to time therefore , c represents time t

$\text{c = t = [ T ]} \\ \\ \boxed{ \: \text{c } = [ M {}^{0} L {}^{0} T ] \: }$

As LHS represents velocity , Each term of RHS must represent velocity.

at = v or a = v/t

$\text{a} = \frac{[ L{T}^{ - 1} ]}{ [{T }]} \\ \\ \boxed{ \: \text{a} = {[ {M }^{0} L{T}^{ - 2} ]} \: }$

And,

b / (t + c) = v

b = v ( t + c )

$\text{b} = [ L{T}^{ - 1} ] ( \: T+ T \: )\\ \\ \text{b} = [ L{T}^{ - 1} ] \: (T)\\ \\ \boxed{ \: \text{b} = [ M {}^{0} L{T}^{ 0} ] \: }$

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Answer 2

Answer:

the above answer is correct...

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