Question

A motor cycle moving with a speed of 5m/s is subjected to an acceleration of 0.2m/s square. calculate the speed of the motor cycle after 10 second and distance travelled in this time​

Answer 1

$\Large{\underline{\underline{\green{\tt{Given}}}}}$

• Initial velocity (u) = 5m/s
• Acceleration (a) = 0.2m/s²
• Time taken (t) = 10 sec

$\Large{\underline{\underline{\green{\tt{Find\;out}}}}}$

• Speed after 10second
• Distance travelled

$\Large{\underline{\underline{\green{\tt{Solution}}}}}$

According to first equation of motion

➨ v = u + at

➨ v = 5 + 0.2 × 10

➨ v = 5 + 2

➨ v = 7m/s

Now, according to third equation of motion

➨ v² = u² + 2as

➨ (7)² = (5)² + 2 × 0.2 × s

➨ 49 = 25 + 0.4s

➨ 49 - 25 = 0.4s

➨ 24 = 0.4s

➨ s = 60m

Hence,

• Speed after 10s = 7m/s
• Distance travelled = 60m

Answer 2

Answer:

• Initial Velocity ( u) = 5 m/s
• Acceleration ( a) = 0.2 m/s²
• Time ( t) = 10 seconds

☢ $\underline{\textsf{By the First Equation of Motion :}}$

$\dashrightarrow\sf\:\:v=u+at\\\\\\\dashrightarrow\sf\:\:v=5\:m/s+(0.2\:m/s^2 \times 10\:s)\\\\\\\dashrightarrow\sf\:\:v = 5\:m/s + 2\:m/s\\\\\\\dashrightarrow\sf\:\:v = 7\:m/s\qquad\bigg\lgroup\bf Final\: Velocity\bigg\rgroup$

$\rule{150}{1.5}$

☢ $\underline{\textsf{By the Third Equation of Motion :}}$

$\dashrightarrow\sf\:\:v^2-u^2=2as\\\\\\\dashrightarrow\sf\:\:(7 \:m/s)^2-(5\:m/s)^2=2 \times 0.2 \:m/s^2 \times s\\\\\\\dashrightarrow\sf\:\: (7\:m/s + 5\:m/s)(7\:m/s - 5\:m/s) = 0.4\:m/s^2 \times s\\\\\\\dashrightarrow\sf\:\:12\:m/s \times 2\:m/s = 0.4\:m/s^2 \times s\\\\\\\dashrightarrow\sf\:\: \frac{12\:m/s \times 2\:m/s}{0.4\:m/s^2} = s\\\\\\\dashrightarrow\sf\:\:3 \times 2\:m \times 10 = s\\\\\\\dashrightarrow\sf\:\:s = 60\:m\qquad\bigg\lgroup\bf Distance\: Travelled\bigg\rgroup$

⠀

$\therefore\:\underline{\textsf{Distance travelled in this time is \textbf{60 m}}}.$