Question

Q.8 A car moves with a velocity
of 12 m/s and stops on
application of break with a de-
acceleration of 6 m/s. Find the
distance travelled by car, if it
stops in next 2 seconds.​

Answer 1

Answer:

• The distance travelled by car is 12m.

Explanation:

Given that,

• Initial velocity (v) = 12 m/s
• Final velocity (u) = 6 m/s
• Time (t) = 2 sec

❤ Applying 1st eqn. of motion, ❤

v = u + at

[ Putting values ]

↪ 6 = 12 + a × 2

↪ 6 - 12 = 2a

↪ -6 = 2a

↪ a = -6/2

↪ a = -3 m/s²

❤ Now, Applying 2nd eqn. of motion, ❤

s = ut + 1/2 × at²

[ Putting values ]

↪ s = 12 × 2 + 1/2 × (-3) × (2)²

↪ s = 24 + 1/2 × -3 × 4

↪ s = 24 + 1/2 × -12

↪ s = 24 - 12

↪ s = 12m.

$\red\bigstar$ Distance travelled = 12m $\green\bigstar$

Answer 2

Explanation:

ᴛʜᴇ ᴅɪsᴛᴀɴᴄᴇ ᴛʀᴀᴠᴇʟʟᴇᴅ ʙʏ ᴛʜᴇ ᴄᴀʀ ɪs :- 12m