A car starts from rest and attains velocity of 20 m/s in 20 sec . The distance traveled in car during this period is
Answers
Answer:
200m
Explanation:
Given:
The initial velocity of the car, u = 0
The final velocity of the car, v = 20 m/s
The total time taken by the car, t = 20 sec
To Find:
The distance travelled by the car during this period.
Calculation:
- Using the first equation of motion, we have:
v = u + at
⇒ 20 = 0 + a × 20
⇒ a = 20 / 20
⇒ a = 1 m/s²
- Now using second equation of motion, we get:
s = ut + 1/2 at²
⇒ s = 0 × 20 + 1/2 × 1 × (20)²
⇒ s = 400/2
⇒ s = 200 m
- So, the distance travelled by the car during this period is 200 m.
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Given :
▪ Initial velocity = zero (i.e. rest)
▪ Final velocity = 20mps
▪ Time interval = 20s
To Find :
▪ Distance covered by car in the given interval of time.
Concept :
✏ Since, acceleration has said to be constant throughout the journey, we can easily apply equation of kinematics to solve this type of questions.
First equation of kinematics :
☆ v = u + at
Third equation of kinematics :
☆ v² - u² = 2as
- s denoted distance
- u denotes initial velocity
- t denotes time
- a denotes acceleration
- v denotes final velocity
Calculation :
☢ Acceleration of car :
→ v = u + at
→ 20 = 0 + a(20)
→ a = 20/20
→ a = 1m/s²
☢ Distance covered by car :
→ v² - u² = 2as
→ (20)² - (0)² = 2(1)s
→ 400 = 2s
→ s = 400/2