Question

Find the initial velocity of a car ,which is stopped in 20 seconds by applying brakes the retardation due to brakes is 4.5 metre per second square?​

Answer 1

Given :-

Time = 20 seconds

Acceleration = 4.5 m/s²

Final velocity = 0 m/s (As it stops)

To Find :-

Initial velocity

Solution :-

We know that

v = u + at

Where

v = final velocity

u = initial velocity

a = acceleration

t = time

0 = u + (-4.5)(20)

0 = u + (-90)

0 - u = -90

- u = -90

u = 90

Therefore

Initial velocity is 90 m/s

Know More :-

Equations of motion

v = u + at

s = ut + 1/2 × at²

v² - u² = 2as

Where

v = final velocity

u = initial velocity

a = acceleration

t = time

s = distance

Answer 2

Provided that:

• Time taken to stop = 20 s
• Final velocity = 0 m/s
• Acceleration = -4.5 m/s sq.

Don't be confused!

• Final velocity cames as zero because the car is stopped as brakes are applied to the car.

• Acceleration cames in negative because the car retards and it retards due to application of brakes and we already know that acceleration is the inverse of retardation and vice versa! Also acceleration is positive and retardation is negative.

To calculate:

• Initial velocity

Solution:

• Initial velocity = 90 mps

Using concept:

• First equation of motion

Using formula:

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

Required solution:

$:\implies \sf v \: = u \: + at \\ \\ :\implies \sf 0 = u + (-4.5)(20) \\ \\ :\implies \sf 0 = u + (-90) \\ \\ :\implies \sf 0 = u - 90 \\ \\ :\implies \sf 0 + 90 \: = u \\ \\ :\implies \sf 90 \: = u \\ \\ :\implies \sf u \: = 90 \: ms^{-1} \\ \\ :\implies \sf Initial \: velocity \: = 90 \: ms^{-1}$

Additional information:

Difference between distance and the displacement is mentioned:

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