Question

A particle’s position as a function of time is

described as

y(t) 2t^2 +3t+ 4

. What is the average velocity of the particle from t =0 to t= 3 sec?

(a) 3 m/s (b) 6 m/s

(c) 9 m/s (d) 12 m/s​

Answer 1

Given :-

 $y(t) = 2t^{2} + 3t + 4$

To find :-

the average velocity of the particle

from t =0 to t = 3 sec

Solution :-

As, we know that

$average velocity = \frac{Total displacement}{Total time}$

as,

$y(t) = 2t^{2} + 3t + 4$

∴ Total displacement ⇒

⇒ Δy = y(3) - y(0)

⇒ $= 2(3^{3} ) + 3 (3) + y -0-0-y$

⇒ ( 18 + 9)m

⇒ 27 m

∴ Average velocity = Δy/Δt = 27/3 m/s = 9 m/s

so our answer is 9m/s

thank you

Answer 2

$\bigstar \underline{\underline{\bf Given:-}}$

• A particle’s position as a function of time is  y(t)=2t²+3t+4.
• Time interval ,t=0to3sec.

$\bigstar \underline{\underline{\bf To\ find:-}}$

• The average velocity of the particle from t =0 to t= 3 sec.

$\bigstar \underline{\underline{\bf Solution:-}}$

We know:

❥ Avg velocity = Total displacement/Total time

Now,

⇒ Total displacement = Δy

Where,

=> Δy = y2 - y1

=> y2 = position of particle at t=3s.

=> y1 = position of particle at t=0

And,

=> y1 =2(0)+3(0)+4

=> y2 =2(3)² +3(3)+4

⇒ Δy = y(3) - y(0)

⇒ (18+9)

⇒ 27m.

❥Avg velocity=Δy/Δt

=> Δt = t2-t1 = 3-0 = 3

Therefore,

=> Avg velocity = 27/3 =9m/s.

Hence,Option(C) is correct .

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