Question

3. An object is moving with an initial velocity of 23 m/s. It is then subject to a constant acceleration of 3.5 m/s2 for 12 s. How far will it have traveled during the time of its acceleration?

Answer 1

Given:

Initial Velocity (u) : 23 m/s

Timec(t) : 12 s

Acceleration (a) : 3.5/m²

To find:

Displacement (s) of the object

Formula used:

• $s = ut + \frac{1}{2} a {t}^{2}$

Solution:

Substituting the value of u, a & t in above equation, we get:

$s = (23 \times 12) + (\frac{1}{2} \times 3.5 \times 12 \times 12)$

$= > s = 276 + 252$

$= > s = 528 \: m$

Hence, the object travels 528 m and that's the answer.

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Answer 2

Given:

• Initial Velocity (u) = 23m/s.

• Acceleration (a) = 3.5m/s².

• Time (t) = 12sec.

To Find:

• Distance (s) = ?

Formula Used:

• s = ut + ½ at².

Solution:

Placing values,

➙ s = (23 × 12) + ½ × 3.5 × (12²)

➙ s = 276 + ½ × 3.5 × 144

➙ s = 276 + 252

➛ s = 528m.

Hence,

• Distance travelled by the object is 528m.

ExtraShorts:

• v = u + at.

• s = ut + ½ at².

• v² = u² + 2as.

Here,

• v = Final Velocity.

• u = Intial Velocity.

• a = Acceleration.

• t = Time.

• s = Distance.