Question

A car is moving with an initial velocity 2v0. It increases its velocity to 4v0 in t seconds. The relation which describes the distance covered in t second (considering uniformly accelerated motion) is

Answer 1

Answer:

4v0 - 2v0 = a t

2v0/t = a

S Or distance = 2v0t + v0t

S = 3v0t

Answer 2

The relation which describes the distance covered in t second (considering uniformly accelerated motion) is 3 v0 × t.

Given: The initial velocity is 2 v0. It increases its velocity to 4 v0 in t seconds.

To Find: The relation which describes the distance covered in t second (considering uniformly accelerated motion) is

Solution:

• We can solve the numerical using the formulas,

         v = u + at                                                      ...(1)

where v = final velocity, u = initial velocity, a = acceleration, t = time.

         v² = u² + 2aS                                                ...(2)

where  v = final velocity, u = initial velocity, a = acceleration, S = distance.

Coming to the numerical, we are given;

  Initial velocity (u) = 2 v0,  Final velocity (v) = 4 v0, time = t

Putting respective values in (1), we get;

   v = u + at  

⇒  4 v0 = 2 v0 + a × t

⇒  a = 2 v0 / t                                                           ...(3)

Putting (3) in (2), we get;

    v² = u² + 2aS        

⇒ ( 4 v0 )² = ( 2 v0 )² + 2 × ( 2 v0 / t  ) × S

⇒ 16 v0²  = 4 v0² + ( 4 v0 / t ) × S

⇒ 12 v0²  = ( 4 v0 / t ) × S

⇒ 3 v0 × t = S

Hence, the relation which describes the distance covered in t second (considering uniformly accelerated motion) is 3 v0 × t.

#SPJ2