A car starts from rest and reaches a velocity of 20m/s in 4s.
(a) Find its acceleration
(b) Calculate the distance covered
show your working
Answers
Explanation:
acceleration= 20-0/4= 5m/s²
Answer- The above question is from the chapter 'Kinematics'.
Some important terms and formulae:-
1. Velocity- It is the displacement per unit time.
S.I. Unit of Velocity- m/s
It is a vector quantity as it possesses magnitude and direction.
2. Acceleration- It is the rate of change of velocity.
S.I. Unit of Acceleration- m/s²
It is also a vector quantity.
Negative acceleration is called retardation.
3. Distance- It is the path length transversed by an object.
S.I. Unit of Distance- m
It is a scalar quantity.
4. Displacement- It is the shortest distance between the initial and final point.
S.I. Unit of Displacement- m
It is a vector quantity.
5. Equations for uniformly accelerated motion-
Let u = Initial velocity of a particle
v = Final velocity of a particle
t = Time taken
s = Distance travelled in the given time
a = Acceleration
1) v = u + at
2) s = at² + ut
3) v² - u² = 2as
6. Average Speed = Total distance ÷ Total time
7. Average Velocity = Total displacement ÷ Total time
Given question: A car starts from rest and reaches a velocity of 20 m/s in 4s.
(a) Find its acceleration
(b) Calculate the distance covered
Show your working.
Answer: Initial Velocity (u) = 0 m/s
Final Velocity (v) = 20 m/s
Time (t) = 4 s
(a) Acceleration (a) = ?
Using 1st equation of motion, v = u + at
20 = 0 + 4a
4a = 20
∴ a = 5 m/s²
(b) Distance (s) = ?
Using 3rd equation of motion, v² - u² = 2as,
20² - 0² = 2 × 5 × s
s = 400 ÷ 10