Question

A motorcycle of mass 20Kg moving with a speed of 10 m/s is subjected to an acceleration of 0.2 m/s2. Calculate a) Speed of the motorcycle after 5 seconds b) Distance travelled in this time c) Force acting on the motorcycle.

Answer 1

Given :

➳ Mass of motorcycle = 20kg

➳ Initial speed = 10m/s

➳ Acceleration = 0.2m/s²

To Find :

⟶ Final speed after 5s.

⟶ Distance travelled in this time.

⟶ Force acting on the motorcycle.

SoluTion :

✴ Final speed :

◕ Acceleration is defined as the rate of change of speed.

⇒ a = (v - u) / t

⇒ 0.2 = (v - 10) / 5

⇒ v - 10 = 0.2 × 5

⇒ v - 10 = 1

⇒ v = 11m/s

✴ Distance travelled in this time :

➨ v² - u² = 2as

➨ 11² - 10² = 2(0.5)s

➨ 121 - 100 = (1)s

➨ s = 21m

✴ Force acting on motorcycle :

➢ Force is defined as the product of mass and acceleration. (F = ma)

➝ F = ma

➝ F = 20×0.2

➝ F = 4N

Answer 2

• Given -

• Mass of motorcycle = 20kg
• Speed (Initial Velocity) = 10m/s
• Acceleration = 0.2 m/s^2
• Time = 5sec

• To Find -

• Final Speed after 5 sec?
• Total Distance travelled in this time period
• Force acted on motorcycle

• Solution -

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★ Final Speed =

We know Formula for acceleration -

$♦a = \frac{v - u}{t}$

Placing Values,

$⟹0.2 = \frac{v - 10}{5}$

$⟹v - 10 = 0.2 \times 5$

$⟹v - 10 = 1$

$⟹v = 1 + 10$

⛬ v = 11 m/s

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★ Total Distance =

We know 3rd eqn of motion -

$</strong><strong>♦</strong><strong> {v}^{2} - {u}^{2} = 2as$

Placing Values,

$⟹ {11}^{2} - {10}^{2} = 2(0.5)s$

$⟹121 - 100 = (1)s$

$⟹s = 121 - 100$

⛬ s = 21m

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★ Force acted on Motorcycle =

We know Formula For Force -

♦ F = ma

Placing Values,

$⟹</strong><strong>F</strong><strong> = 20 \times 0.2$

⛬ F = 4N

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