Question

A particle start with an initial velocity 2.5 m/s along thepositive x direction and it accelerate uniformly at rate 0.50 m/s2 find the distance travelledby it in the first two second

Answer 1

Explanation:

We have,

x=ut+12at2

=(2.5ms)(2)+12(0.50ms2)(2s)2

=5.0m+1.0m=6.0m.

Since the particle does not turn back it is also the distnce travelled.

Answer 2

$\maltese$ Answer $\maltese$

Given:

⇒ Initial Velocity (u) = 2.5 m/s

⇒ Acceleration (a) = 0.50 m/s²

⇒ Time (t) = 2 seconds

To Find:

The Distance (S) travelled by it in the first two second

Solution:

We know that,

$\red{\underline{\large{\rm {Equation \ of \ Motion}}}}$

$\green{\boxed{\boxed{\rm S=ut+\dfrac{1}{2}at^{2}}}}$

According to the Question,

We are asked to find the Distance (S) travelled by it in the first two second  along the positive x-direction.

So, Using the Formula,

Given that,

• u = 2.5 m/s
• a = 0.50 m/s²
• t = 2 seconds

Therefore,

Substituting the above values in the Formula,

We get,

$\orange{\rm \implies S=(2.5)(2)+\dfrac{1}{2} (0.50)(2)^{2} \ \ m}$

$\blue{\rm \implies S=5+\dfrac{1}{2} (0.50)(4) \ \ m}$

$\green{\rm \implies S=5+\dfrac{1}{2}(2) \ \ m}$

$\orange{\rm \implies S=5+1 \ \ m}$

$\pink{\rm \implies S=6 \ \ m}$

$\red{\large{\boxed{\boxed{\rm Distance \ Travelled \; (S)=6 \ m}}}}$

Hence,

The Distance (S) travelled in the first two second  = 6 m

Distance (S) = 6 m