Question

in 1.5s the speed of the jumper increases from zero to 10.5m/s calculate her average acceleration during this time​

Answer 1

Acceleration = ∆v/∆t

→ 10.5/1.5

→ 105/15

→ 7 m/s²

Answer 2

Answer :

›»› The Acceleration of jumper = 7 m/s²

Given :

• Initial velocity of jumper (u) = 0 m/s
• Final velocity of jumper (v) = 10.5 m/s
• Time taken by jumper (t) = 1.5 sec

To Find :

• Acceleration of jumper (a) = ?

Required Solution :

→ Initial velocity (u) = 0 m/s (because jumper increase from zero)

† From first equation of motion

⇒ v = u + at

⇒ 10.5 = 0 + a × 1.5

⇒ 10.5 = 0 + 1.5a

⇒ 10.5 = 1.5a

⇒ a = 10.5/1.5

⇒ a = 7 m/s²

║Hence, the Acceleration of jumper is 7 m/s².║

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Additional Information :

★ First equation of motion :

⪼ v = u + at

Where,

• v is the Final velocity in m/s.
• u is the Initial velocity in m/s.
• a is the Acceleration in m/s².
• t is the Time taken in second.

★ Second equation of motion :

⪼ s = ut + ½ at²

Where,

• s is the Distance travelled in m.
• u is the Initial velocity in m/s.
• a is the Acceleration in m/s².

★ Third equation of motion :

⪼ v² = u² + 2as

Where,

• v is the Final velocity in m/s.
• u is the Initial velocity in m/s.
• a is the Acceleration in m/s².
• s is the Distance travelled in m.