Question

A car is travelling at 36km/hr speeds up to 72km/hr in 5 seconds. Find its acceleration. If the same car stops in 20 seconds what is the retardation...
Plz say quick.
With proper process...

Answer 1

WHAT IS ACCELERATION?

The rate at which the velocity of a body changes per unit time is called Acceleration. Its unit is m/sec^2.

WHAT IS RETARDATION?

The negative acceleration is known as retardation.

It happens when the final velocity is less than the initial velocity.

FORMULA :-

Both acceleration and retardation has same formula.

$a = \frac{(v - u)}{t}$
where,

a = acceleration/retardation

v = final velocity

u = initial velocity

t = time taken

YOUR ANSWER :-

GIVEN :-

CASE 01                        CASE 02

v =72km/hr=20m/s v=0km/hr=0m/sec

u=36km/h=10m/s u=36km/hr=10m/s

t = 5 seconds   t = 20 seconds

SOLUTION :-

For Case 01

$a = \frac{(v - u)}{t} \\ \\ = \frac{(20 - 10)}{5} \\ \\ = \frac{10}{5} \\ \\ = 2msec^{-2}$

For Case 02

$a=\frac{(v - u)}{t} \\\\\\ \implies a = \frac{(0 - 10)}{20}\\\\\\ \implies a = \frac{-10}{20} \\\\\\ \implies a = - 0.5\: m\, sec^{ - 2}$

......................................................................

Answer 2

Intial Speed of Car(u)=36 km/h

In m/s=36×5/18=10 m/s

Final Velocity (v)=72 km/h

In m/s=72×5/18

=20 m/s

So,

Intial Velocity (u)=10 m/s.

Final Velocity (v)=20 m/s

Time(t)=5 seconds

Use Equation of Motion

v=u+at

20=10+a×5

20-10=5a

5a=10

a=10/5

a=2

Acceleration=2 m/s²

Now In second Case

Intial Velocity (u)=72 km/h=20 m/s

Final Velocity after Applying Break when it Stop(v)=0 m/s

Time Taken(t)=20 seconds

Use Equation of Motion

v=u+at

0=20+a×20

-2/=20a

a=-2//20

a=-1 m/s²

(1) Acceleration=2m/s²

(2)Retardation=-1 m/s²