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Katarina [22]
3 years ago
8

( In 1.5s the speed of the jumper increases from zero to 10.5 m/s. Calculate her average acceleration during this time.​

Physics
1 answer:
Bas_tet [7]3 years ago
4 0

Answer:

answer 7m/s^2

Explanation:

›»› 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.║

\:

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.

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