Question

Question 2/15 (2 p.)
The displacement of a particle is given by
x=(t - 2)2, where x is in meter and tin
seconds. The distance covered by the
particle in first 4 seconds is :​

Answer 1

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 = $\frac{1}{2}$ at² + ut

3) v² - u² = 2as

6. Average Speed = Total distance ÷ Total time

7. Average Velocity = Total displacement ÷ Total time

8. Some basic formulae using calculus method:

1) Velocity = $\frac{dx}{dt}$

2) Acceleration = $\frac{dv}{dt}$

3) Acceleration (when x is given) = $v \frac{dx}{dt}$

Given question: The displacement of a particle is given by x = (t - 2)² , where x is in meter and t in seconds. The distance covered by the particle in first 4 seconds is:

Solution: x = (t - 2)²

x = t² + 4 - 4t

At t = 0 seconds, x = 0 + 4 - 0 = 4 m

At t = 2 seconds, x = 4 + 4 - 8 = 0 m

At t = 4 seconds, x = 16 + 4 - 16 = 4 m

Distance = 4 + 0 + 4 = 8 m

∴ The distance covered by the particle in first 4 seconds is 8 m.