Question

an object undergoes an acceleration of 10 ms-2 starting from rest. find the distance traveled in 5 seconds.
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Answer 1

✬ Distance = 125 m ✬

Explanation:

Given:

• An object starts from rest.
• Acceleration produced by object is 10 m/s².
• Time given is 5 seconds.

To Find:

• Distance travelled by the object after in given time ?

Solution: Since object started from rest therefore u is 0. Now to find distance we will use third law of Newton.

★ S = ut + 1/2at² ★

• s = distance

• u = initial position = 0 m/s

• a = acceleration = 10 m/s²

• t = time = 5 seconds

Putting all these value in above formula.

$\implies{\rm }$ s = 0 × 5 + 1/2 × 10 × 5 × 5

$\implies{\rm }$ s = 5 × 25

$\implies{\rm }$ s = 125 m

Hence, the distance covered by object in 5 seconds is 125 m.

Answer 2

Answer:

Distance travelled in 5 seconds is 125 m.

Explanation:

Given :-

An object undergoes an acceleration 10 m/s starting from rest.

To find :-

Distance travelled in 5 seconds.

Solution :-

From equation of motion,

S = ut + 1/2at²

Where,

S - Displacement

u - initial velocity

a - acceleration

t - time taken

u will be 0 since the body accelerated from rest.

Substituting we get,

S = 0 × 5 + 1/2 × 10 × 5²

S = 0 + 5 × 25

S = 125

∴ Distance travelled in 5 seconds is 125 m.