Question

A horse starts from rest and accelerates at 3 m/s2

for 20 seconds; calculate

(i) The final velocity.

(ii) The distance travelled.​

Answer 1

GIVEN :-

• Initial velocity ( u ) = 0 m/s.
• Acceleration ( a ) = 3 m/s².
• Time taken , ( t ) = 20 s.

TO FIND :-

• The Final velocity ( v ).
• Distance travelled ( s ).

SOLUTION :-

By using the first equation of motion,

→ v = u + at.

Where,

• v stands for Final velocity
• u stand for Initial velocity
• t stands for Time
• a stands for Acceleration

On Substituting the values we get,

→ v = 0 + 3 × 20

→ v = 0 + 60

→ v = 60 m/s.

.°. The final velocity is 60 m/s.

By using the second equation of motion,

→ s = ut + ½ at².

• s stands for Distance travelled.

On substituting the values we get,

→ s = 0 × 30 + ½ × 3 × (20)²

→ s = 0 + ½ × 3 × 400

→ s = 0 + ½ × 1200

→ s = 0 + 600

→ s = 600 m.

.°. The distance travelled is 600 m.

_______________________________

Newton's first equation of motion :

• v = u + at

Newton's second equation of motion :

• s = ut + ½ at²

Newton's third equation of motion :

• v² = u² + 2as

Answer 2

$\Large\purple{\mathcal{GIVEN}} \large{\pink{\begin{cases} \red{\mapsto \sf Acceleration, a = {3m/s}^2 } \\ \\ \\ \orange{\mapsto \sf Time, t = 20sec } \end{cases}}}$

$\Large\purple{\mathcal{Find}} \large{\red{\begin{cases} \pink{\mapsto \sf Final \: Velocity, v } \\ \\ \\ \blue{\mapsto \sf Distance \: Travelled, s } \end{cases}}}$

$\Large\purple{\mathcal{Solution}}$

Here, horse starts from rest so,

it's initial Velocity, u = 0m/s

Now, we know that

$\underline{\boxed{ \sf v = u + at}} \qquad \quad \lgroup{ \sf {1}^{st} \: eq. \: of \: motion} \rgroup$

where,

• Initial velocity, u = 0m/s
• Acceleration, a = 3m/s²
• Time, t = 20sec

So,

➤ v = u + at

➧ v = (0) + (3)(20)

➧ v = 0 + 60

➧ v = 60m/s

➟ v = 60m/s $\red\bigstar$

____________________________

we, know that

$\underline{\boxed{ \sf s = ut + \dfrac{1}{2} a {t}^{2}}} \qquad \quad \lgroup{ \sf {2}^{nd} \: eq. \: of \: motion} \rgroup$

where,

• Initial Velocity, u = 0m/s
• Time, t = 20sec
• Acceleration, a = 3m/s²

So,

⇒s = ut + $\dfrac{1}{2}$ at²

➲ $\sf s = (0)(20) + \dfrac{1}{2} \times (3) \times {(20)}^{2}$

➲ $\sf s = \dfrac{1}{2} \times (3) \times (400)$

➲ $\sf s = \dfrac{1}{2} \times (1200)$

➲ $\sf s = 600m$

➯ s = 600m $\pink\bigstar$

___________________________

Hence,

• Final Velocity, u = 60m/s
• Distance Travelled, s = 600m