Question

If s = 3t^2 + 4t +9 find
(i) velocity at t = 2s
(ii) avg avelocity between 1 and 4sec​

Answer 1

Given :

s = 3t² + 4t + 9

To find :

(i) velocity at t = 2s

(ii) avg avelocity between 1 and 4secm

Solution :

(i) The velocity of a particle at a particular instant of time is called it's instantaneous velocity.

$\to \sf v_{inst} = \dfrac{ds}{dt}$

$\to \sf v_{inst} = \dfrac{d(3 {t }^{2}+ 4t + 9) }{dt}$

$\to \sf v_{inst} = 6t + 4$

we have to find velocity at 2sec.

$\to \sf v_{inst} = 6(2) + 4$

$\to \sf v_{inst} = 16m/s$

thus, velocity at 2sec is 16m/s.

(ii) For a particle in motion, the ratio of total displacement to the total time interval is called average velocity.

first we have to find displacement at t = 1 and t = 4

displacement in 4sec,

s₄ = 3(4)² + 4(4) + 9

s₄ = 3 × 16 + 16 + 9

s₄ = 48 + 16 + 9

s₄ = 73m

displacement in 1sec,

s₁ = 3(1)² + 4(1) + 9

s₁ = 3 + 4 + 9

s₁ = 16m

we know,

$\sf Average \: velocity = \dfrac{x_2 - x_1}{\Delta t} =\dfrac{\Delta x}{\Delta t}$

According to Question,

$\to \sf v_{avg} = \dfrac{s_4 - s_1}{4 - 1}$

$\to \sf v_{avg} = \dfrac{73 - 16}{3}$

$\to \sf v_{avg} = \dfrac{57}{3}$

$\to \sf v_{avg} = 19m/s$

thus, average velocity between 1 and 4sec is 19m/s.

Answer 2

${\huge{\underline{\underline{\rm{Question:-}}}}}$

Q) If s = 3t² + 4t + 9 , find

(i) velocity at t = 2s

(ii) average velocity between 1 & 4 second .

${\huge{\underbrace{\rm{Answer\checkmark}}}}$

$\boxed{ \sf \: s = 3 {t}^{2} + 4t + 9 \: }$

As you can see , we have the Equation of distance .

(i) Velocity at t = 2s

We know that ,

Velocity = Rate of change of distance w.r.t. time .

So ,

$\rm \: velocity _{inst} = \frac{ds}{dt}$

$\mapsto \rm \: v _{inst} = \frac{d(3 {t}^{2} + 4t + 9)}{dt}$

$\mapsto \rm \: v _{inst} = \frac{d(3 {t}^{2}) }{dt} + \frac{d(4t)}{dt} + \frac{d(9)}{dt}$

$\mapsto \rm \: v _{inst} = 6t + 4 + 0$

$\leadsto \underline{\underline{\rm \: v _ {inst} = 6t + 4 \: }}$

Now , as we are supposed to find velocity at time = 2 sec .

put it in the above equation ..

$\mapsto \rm \: v _{2 \: sec} = 6(2) + 4 \: \frac{m}{ {s}}$

$\mapsto \rm \: v _{2 \: sec} = 12 + 4 \: \frac{m}{ {s} }$

$\longrightarrow \pink{\boxed{ \rm \: v _{2 \: sec} = 16 \: \frac{m}{s} }}$

_________________________

(ii) Average Velocity between 1 & 4 sec .

we have

• $\sf \: t _{1} = 1 \: sec$
• $\sf \: t _{2} = 4 \: sec$

So , Finding the position of the particle at the respective times .

At t = 1 ,

$\mapsto \rm \: s _{1} = 3 {(1)}^{2} + 4(1) + 9$

$\mapsto \rm \: s_{1} = 3 + 4 + 9 \: m$

$\leadsto \underline{\underline{\rm \: s _{1} = 16 \: m \: }}$

At t = 4 ,

$\mapsto \rm \: s _{2} = 3 {(4)}^{2} + 4(4) + 9$

$\mapsto \rm \: s _{2} = 3 \times 16 + 16 + 9 \: m$

$\mapsto \rm \: s _{2} = 48 + 25 \: m$

$\leadsto \rm \underline{ \underline{ \: s _{2} = 73m \: }}$

Now , using the Formula ↓↓

$\boxed{ \sf \: average \: velocity = \frac{ \triangle{x}}{ \triangle{t}} }$

here ,

• ∆x = change in position . &
• ∆t = change in time .

$\mapsto \rm \: average \: velocity = \frac{s _{2} - s _{1} }{t _{2} - t _{1} }$

$\mapsto \rm \: average \: velocity = \frac{73 - 16}{4 - 1} \frac{m}{s}$

$\mapsto \rm \: average \: velocity = \cancel \frac{57}{3} \: \frac{m}{s}$

$\longrightarrow \purple{\boxed{ \rm \: average \: velocity = 19 \: \frac{m}{s} }}$

_________________________

So , Final answer :

1. 16 m/s
2. 19 m/s

• The unit of velocity is always in m/s ( meter per second )