A car starting from rest attains a velocity of 30ms-1 in 5 seconds find(a) acceleration
(b) the distance travelled by the car in 5 seconds
Answers
Required Answer :
Given :
• Initial velocity of car (u) = 0 [ As it starts from rest.]
• Final velocity (v) = 30 m/s
• Time taken (t) = 5 seconds
To calculate :
• Acceleration (a)
• The distance travelled by the car (s)
Calculation :
→ Calculating acceleration -
By using the first equation of motion :
→ v = u + at
Where,
• v = final velocity
• u = initial velocity
• a = acceleration
• t = time taken
Inserting values, we get :
→ v = u + at
→ 30 = 0 + ( a × 5 )
→ 30 = 0 + 5a
→ 5a = 30 - 0
→ 5a = 30
→ a = 30/5
→ a = 6 m/s² [Acceleration]
Therefore, acceleration of the car is 6 m/s² .
→ Calculating distance travelled -
By using the third equation of motion :
→ s = ut + ½ at²
Where,
• v = final velocity
• u = initial velocity
• a = acceleration
• s = distance travelled
Inserting values, we get :
→ s = ut + ½ at²
→ s = ( 0 × 5 ) + ½ × 6 × (5)²
→ s = 0 + ½ × 6 × 25
→ s = 1 × 3 × 25
→ s = 75 m [Distance]
Therefore, distance travelled by the car in 5 seconds is 75 m.
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Extra Information :
ꜰɪʀꜱᴛ ᴇQᴜᴀᴛɪᴏɴ ᴏꜰ ᴍᴏᴛɪᴏɴ :
- v = u + at
→ v = final velocity
→ u = initial velocity
→ a = acceleration
→ t = time taken
ꜱᴇᴄᴏɴᴅ ᴇQᴜᴀᴛɪᴏɴ ᴏꜰ ᴍᴏᴛɪᴏɴ :
- s = ut + ½ at²
Where,
→ v = final velocity
→ u = initial velocity
→ a = acceleration
→ s = distance travelled
ᴛʜɪʀᴅ ᴇQᴜᴀᴛɪᴏɴ ᴏꜰ ᴍᴏᴛɪᴏɴ :
- v² - u² = 2as
Where,
→ v = final velocity
→ u = final velocity
→ a = acceleration
→ s = distance travelled
- When the body starts from rest, its initial velocity is 0.
- When the body comes to rest or apply breaks, its final velocity is 0.
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