Question

A car accelerates uniformly from rest and attains a velocity of

40m/s in 10 sec. Calculate (i) acceleration (ii)distance covered by

the car.​

Answer 1

Answer:

ᴛɪᴍᴇ (ᴛ) = sᴇᴄᴏɴᴅs

ᴀᴄᴄᴇʟᴇʀᴀʀᴛɪᴏɴ (ᴀ) = . ᴍ/s

ᴡᴇ ᴋɴᴏᴡ ᴠ = ᴜ + ᴀᴛ

⇒ ᴍ/s = ᴜ + .

s

ᴍ

⇒ ᴍ/s = ᴜ +

s

ᴍ

⇒ ᴜ = ( - ) ᴍ/s

= ᴍ/s

∴ ɪɴɪᴛɪᴀʟ ᴠᴇʟᴏᴄɪᴛʏ ᴏғ ᴛʜᴇ ᴄᴀʀ = ᴍ/s

Answer 2

Answer:

• Acceleration is 4 m/s².
• Distance travelled is 200 m.

Given:

• Final velocity = 40 m/s
• Time taken = 10 sec

Explanation:

$\rule{300}{1.5}$

Here, we are given that a car accelerates uniformly from rest and attains a velocity of 40 m/s in 10 sec. We are asked to find it's acceleration and the distance travelled by it.

Since, it starts from rest, so it's initial velocity will be zero.

From first kinematic equation we know,

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

Substitute the values,

$\longrightarrow{\sf{40\ =\ 0\ +\ a \times 10}} \\ \\ \\ \longrightarrow{\sf{40\ =\ 0\ +\ 10a}} \\ \\ \\ \longrightarrow{\sf{\dfrac{40}{10}\ =\ a}} \\ \\ \longrightarrow{\sf{4\ =\ a}} \\ \\ \\ \longrightarrow{\sf{a\ =\ 4\ m/s^2}}$

∴ The acceleration of the car is 4 m/s².

$\rule{300}{1.5}$

Now, let us find out the distance travelled by the car. From second kinematic equation we know,

$\longrightarrow{\sf{s\ =\ ut\ +\ \dfrac{1}{2}at^2}}$

Substitute the values,

$\longrightarrow{\sf{s\ =\ 0 \times 10\ +\ \dfrac{1}{2} \times 4 \times (10)^2}} \\ \\ \\ \longrightarrow{\sf{s\ =\ 0\ +\ \dfrac{1}{2} \times 4 \times 10 \times 10}} \\ \\ \\ \longrightarrow{\sf{s\ =\ \dfrac{1}{2} \times 4 \times 100}} \\ \\ \\ \longrightarrow{\sf{s\ =\ \dfrac{1}{\cancel2} \times \cancel4 \times 100}} \\ \\ \\ \longrightarrow{\sf{s\ =\ 2 \times 100}} \\ \\ \\ \longrightarrow{\sf{s\ =\ 200\ m}}$

∴ The distance travelled by the car is 200 m.

$\rule{300}{1.5}$