Question

If a sports car can go from rest to 27 m s -1 in 9.0 second , what is the magnitude of its average acceleration?

Answer 1

Correct question :

If a sports car can go from rest to 27 m/s in 9.0 seconds, what is the magnitude of its acceleration?

Solution

Here we have,

✪ Initial speed = 0 [It starts from rest]

✪ Final speed = 27 m/s

✪ Time taken = 9 seconds.

✪ Acceleration = ?

We will use 1st equation of motion:-

✪ v = u + at

⇒ 27 = 0 + a × 9

⇒ 27 = 9a

⇒ a = 27/9

⇒ a = 3 m/s²

∴Magnitude of acceleration = 3 m/s²

✪ Equations of motion :-

1. v = u + at
2. s = ut + ½ at²
3. v² - u² = 2as

Answer 2

Given:-

Initial velocity of car = 0 [ start from rest ]

Final velocity= 27m/s

Time taken = 9 seconds

To find :-

The magnitude of average acceleration or acceleration of car

Magnitude is a numerical value !

Now by using 1st equation of motion

$\boxed{\sf { v= u+at}}$

$\longrightarrow\sf 27= 0+a\times 9\\ \\ \longrightarrow\sf 27=9a\\ \\ \longrightarrow\sf \cancel{\dfrac{27}{9}}=a\\ \\ \longrightarrow\sf a= 3m/s^2$

$\underline{\bigstar{\sf\ \ Magnitude\ of \ Acceleration= 3m/s^2}}$

Some other equation of motion !

$\bullet\sf s= ut+\dfrac{1}{2}at^2\\ \\ \bullet\sf v^2=u^2+2as$

Where ,

•s= Distance

•v= final velocity

•u= Initial velocity

•t= time

•a= acceleration