Question

If a car starts from rest with an acceleration of 2 ms^-2, the distance travelled by car after 2 sec would be
1. 4 m
2. 2 m
3. 8 m
4. 6 m

Answer 1

Correct Answer - Option 1 : 4 m

Concept:

Equations of Motion

• The equations of motion establish the relationship between acceleration, time, distance, initial speed, and final speed for a body moving in a straight line with uniform acceleration.
• This is for constant acceleration. 

The equations are

v = u + at - (1)

\(s = ut + \frac{1}{2}at^2\) - (2)

v2 = u2 + 2as - (3)

v is final velocity, u  is initial velocity, t is time, a is acceleration, s is the distance travelled. 

Calculation:

Given

initial speed u = 0 

final speed v = ?

acceleration a =2 m s ^{- 1}. 

distance s

\(s = ut + \frac{1}{2}at^2\)

\(\implies s = (0) \times 2 + \frac{1}{2} \times 2\times (2)^2\)

⇒ s = 4 m

So the correct option is 4 m.

• Displacement: The shortest distance between two points is called displacement.
• Velocity: The rate of change of displacement is called velocity

v = ds / dt

• Acceleration: The rate of change of velocity is called acceleration

a = d2 s / dt