Question

A racing car has a uniform acceleration of 2m/sec square.what distance will it covers in 8sec after start

Answer 1

Answer :-

Given :-

• Acceleration, a = 2 m/s²
• Initial velocity, u = 0 m/s
• Time taken, t = 8sec

To find :-

• Distance - s

Solution :-

• a = 2 m/s²
• u = 0 m/s
• t = 8sec

Substituting the value in 2nd equation of motion :-

$\sf s = ut + 1/2 at^2$

$\sf s = 0 + \frac{1}{2} \times 2 \times 8^2$

$\sf s = \frac{1}{\cancel 2} \times \cancel 2 \times 8^2$

$\sf s = 8 \times 8$

$\sf s = 64 m$

Answer 2

Given :-

• Acceleration (a) = 2 m/s²
• Initial Velocity (u) = 0 m/s (as it is starting from rest)
• Time taken (t) = 8 s

To Find :-

• Distance Covered

Here, in this question it's saying that a racing car started from rest, so here initial Velocity is 0 and it's acceleration is 2 m/s² and if it travels 8s , what would be the distance after that.

So, here a , u , t is given

So, by the formula s = ut + ½at²

We can get the distance

Solving Further

→ s = 0 × 8 + ½ × 2 × 8²

→ s = 0 + ½ ×2 × 64

→ s = 1 × 64

→ s = 64

Thus, the distance is 64 m.