Question

The initial speed of a particle is 3 m/s and it is accelerating with a constant acceleration of 2 m/s². Find

(a) speed of particle after 5 seconds.

(b) distance travelled in 5 seconds.​

Answer 1

Provided that:

• Time taken = 5 seconds
• Initial velocity = 3 m/s
• Acceleration = 2 m/s sq.

To calculate:

• Speed after 5 seconds
• Distance after 5 seconds

Solution:

• Speed after 5 seconds = 13 m/s
• Distance after 5 seconds = 40 m

Using concepts:

• First equation of motion
• Third equation of motion

Using formulas:

• First equation of motion is given by,

• ${\small{\underline{\boxed{\sf{v \: = u \: + at}}}}}$

• Second equation of motion is given by,

• ${\small{\underline{\boxed{\sf{s \: = ut \: + \dfrac{1}{2} \: at^2}}}}}$

Where, v denotes final velocity, u denotes initial velocity, a denotes acceleration, s denotes displacement or distance or height, t denotes time taken

Required solution:

~ Firstly let us calculate final velocity that is speed of particle after 5 seconds.

→ v = u + at

→ v = 3 + 2(5)

→ v = 3 + 10

→ v = 13 m/s

→ Final velocity = 13 m/s

~ Now let's calculate the distance!

→ s = ut + ½ at²

→ s = 3(5) + ½ × 2 × (5)²

→ s = 3(5) + ½ × 2 × 25

→ s = 15 + ½ × 2 × 25

→ s = 15 + ½ × 50

→ s = 15 + 1 × 25

→ s = 15 + 25

→ s = 40 m

→ Distance = 40 metres

Additional information:

$\begin{gathered}\boxed{\begin{array}{c|cc}\bf Distance&\bf Displacement\\\frac{\qquad \qquad \qquad\qquad\qquad \qquad\qquad\qquad}{}&\frac{\qquad \qquad \qquad\qquad\qquad \qquad\qquad\qquad}{}\\\sf Path \: of \: length \: from \: which &\sf The \: shortest \: distance \: between \\ \sf \: object \: is \: travelling \: called \: distance. &\sf \: the \: initial \: point \: \& \: final \\ &\sf point \: is \: called \: displacement. \\\\\sf It \: is \: scalar \: quantity. &\sf It \: is \: vector \: quantity \\\\\sf It \: is \: positive \: always &\sf It \: can \: be \: \pm \: \& \: 0 \: too \end{array}}\end{gathered}$