Question

A car starts with a velocity of 10m/s and accelerates at the rate of 5m/s2 the final velocity when it travels a distance of 30m is:
(a) 400m/s
(b) 20m/s (c)
200m/s
(d) 150m/s

Answer 1

Given :

• Initial velocoty of the car = 10 m/s
• Acceleration of car = 5 m/s²
• Distance = 30 m

To Find :

• The Final velocity of the car

Solution :

Using third equation of motion ,

$\\ \star \: {\boxed{\sf{\purple{ {v}^{2} - {u}^{2} = 2as }}}}$

Where ,

• v is final velocity
• u is initial velocity
• a is acceleration
• s is distance

We have ,

• u = 10 m/s
• v = (let v)
• s = 30 m
• a = 5 m/s²

Substituting the values ,

$\\ : \implies \sf \: {v}^{2} - {(10)}^{2} = 2(5)(30)$

$\\ : \implies \sf \: {v}^{2} - 100 = 300$

$\\ : \implies \sf \: {v}^{2} = 300 + 100$

$\\ : \implies \sf \: {v}^{2} = 400$

$\\ : \implies \sf \: v = \sqrt{400}$

$\\ : \implies{\underline{\boxed{\pink{\mathfrak{ v = 20 \: m {s}^{ - 1} }}}}} \: \bigstar$

Hence ,

• The Final velocity of car satisfying the given conditions is 20 m/s

Answer 2

A N S W E R :

• Option (b) is correct.
• 20m/s

Given :-

• Initial Velocity of the car = 10m/s
• Acceleration of car = 5m/s²
• Distance = 30m

To find :-

• Find The Velocity of the car ?

★ $\large\underline{\frak{As~we~know~that,}}$

$\large\dag$ Formula Used :

• $\boxed{\bf{v² - u² = 2as}}$$\large\star$

Solution :-

• Third equation of motion

So,

• v is Final velocity
• u is Initial velocity
• a is Acceleration
• s is Distance

Now,

• u = 10m/s
• v = ( v )
• s = 30m
• a = 5m/s²

• Substituting the value,

:$\implies$v² - (10) = 2(5) (30)

$~~~~~$:$\implies$v² - 100 = 300

$~~~~~~~~~~$:$\implies$v² = 300 + 100

$~~~~~~~~~~~~~~~$:$\implies$v² = 400

$~~~~~~~~~~~~~~~~~~~~$:$\implies$v = ${\sf{\sqrt{400}}}$

$~~~~~~~~~~~~~~~~~~~~~~~~~$:$\implies$${\underline{\boxed{\pink{\frak{v~= {20ms}^{-1}}}}}}$★

$\large\dag$ Hence,

• The final velocity of car satisfying the given condition is = $\large\underline{\rm{20ms}}$