Question

when a car driver travelling at an initial velocity of 10 m/s applies brakes and beings the car of rest in 5 second, what will be its asseleration​

Answer 1

Given :-

• The initial velocity (u) = 10m/s
• The final velocity (v) = 0m/s
• Time taken by car (T) = 5 sec

To Find :-

• The acceleration of car = ?

Solution :-

• To calculate the acceleration at first we have to focus on the given Question. As given in the question that a car driver travelling at an initial velocity of 10 m/s applies brakes and beings the car of rest in 5 second. Simply applying formula to calculate the acceleration or restardation of the car.

Formula used :-

⇢ V = U + AT

• V = 0m/s. U = 10m/s. T = 5 sec

⇢ V - U = AT

⇢ A = (V - U)/T

⇢ A = (0 - 10)/5

⇢ A = -10/5

⇢ A = -2m/s²

Hence,

• The acceleration or restardation = -2m/s²

Answer 2

G I V E N :

When a car driver travelling at an initial velocity of 10 m/s applies brakes and beings the car of rest in 5 second, what will be its acceleration.

S O L U T I O N :

GivEn Data ⤵

• Initial Velocity (u) = 10 m/s

• Final Velocity (v) = 0 m/s

• Time (t) = 5 seconds

Finding Acceleration ⤵

To find acceleration we need to use the primary formula,i.e,

$\star\: \underline{ \boxed{ \frak\purple{{ \purple{v = u + at}}}}}$

Putting all the values we get

$\twoheadrightarrow \frak{0=10+a(5) }$

Transposing 10 we get -10

$\twoheadrightarrow \frak{0-10 = 5a}$

$\twoheadrightarrow \frak{5a=-10}$

$\twoheadrightarrow \frak{a=-\frac{10}{5}}$

$\twoheadrightarrow \frak{a=-\cancel\frac{10}{5}}$

$\twoheadrightarrow \boxed{ \frak{ \green{a = - 2}}}$

• Henceforth, the acceleration is -2 m/s²

E X T R A S :

Acceleration (a) : rate of change(increase) of velocity.

Retardation or negative acceleration is the rate of change of decrease in velocity

• It is a vector quantity.

• It can be positive, negative or zero.

Type : Uniform and variable

Let a body is moving with velocity 'u' when a force a a is applied its velocity becomes 'v' in time 't'

$\leadsto \boxed{ \frak{ \green{acceleration(a) = \frac{ v-u}{t}}}}$

Then,

$\leadsto \frak{at=v-u}$

$\leadsto \frak{at+u=v}$

$\star\: \underline{ \boxed{ \frak{ \purple{v = u + at}-1st \:equation \:of\: motion}}}$

$\leadsto \frak{Distance=average velocity×time}$

$\leadsto \frak{S = (\frac{v+u}{2})×t}$

Now, as we got v = u + at that we'll put here

$\leadsto \frak{S = (\frac{u + at + u}{2})×t}$

$\leadsto \frak{S = (\frac{2u + at}{2})×t}$

$\star\: \underline{\boxed{\frak{\green{S = ut + \frac{1}{2}at}}-\frak{2nd \:equation \:of \:motion}}}$

Again,

$\leadsto \frak{S = (\frac{v+u}{2})×t}$

$\leadsto \frak{S = (\frac{v+u}{2})×(\frac{v-u}{a})}$

$\leadsto \frak{S = 2aS =v²-u²}$

$\star \: \underline{\boxed{\frak{\red{S = v² = u²+2aS}}-\frak{3rd\:equstion\:of\: motion}}}$