Question

A taxi travelling at 15 m/s. It driver accelerates with acceleration 3m/s² for 4 s. What is its new velocity?

Answer 1

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A taxi travelling at 15 m/s. It driver accelerates with acceleration 3m/s² for 4 s. What is its new velocity?

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• Initial Velocity (u) = 15 m/s
• Acceleration (a) = 3 m/s²
• Time Interval (t) = 4 sec

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We have to find Final Velocity :-

Use formula :-

$\LARGE \implies{\boxed{\boxed{\green{\tt{a \; = \; \frac{v \; - \; u}{t}}}}}}$

Let final velocity be "v"

(Putting Values)

⇒ 3 = 15 - v/4

⇒ 3 * 4 = 15 - v

⇒ 12 = 15 - v

⇒ 12 - 15 = -v

⇒ -3 = -v

⇒ v  = 3

$\Large{\boxed{\red{\sf{Final \; Velocity \; (v) \; = \; 3 \; m s^{-1}}}}}$

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$\LARGE{\underline{\underline{\sf{Additional \: Information}}}}$

• Acceleration is defined as rate of change of velocity.
• velocity is displacement per time.
• Initial velocity is velocity which is of starting point
• Final velocity is velocity in last point.

S.I  Units

• Acceleration (a) = m/s²
• Velocity = m/s
• Displacement = m

Answer 2

The new velocity of taxi is 27 m/s .

Explanation:

Initial velocity of taxi, u=15 m/s

Acceleration of taxi, a=$3 m/s^2$

Time=4 s

We know that

Acceleration,a=$\frac{v-u}{t}$

Where

v=Final velocity of object

u=Initial velocity of object

t=Time (in sec)

Substitute the values then we get

$3=\frac{v-15}{4}$

$v-15=12$

$v=12+15=27 m/s$

Hence, the new velocity of taxi is 27 m/s .

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