Question

A car is traveling along the road at 8$m{s}^{-1}$. It accelerates at 1 $m{s}^{-2}$ for a distance of 18 m. How fast is it then traveling?

Answer 1

Step 1: Given that:

The initial velocity of the car(u) = $8m{s}^{-1}$

Acceleration in the car(a) = $1m{s}^{-2}$

Distance(s) = $18m$

Step 2: Calculation of the final velocity of the car:

Using the third equation of motion;

$v^2=u^2+2as$

Where; u = initial velocity, v= final velocity and a= acceleration and s= distance covered

Putting the given values, we get

$v^2={\left(8m{s}^{-1}\right)}^2+2\times 1m{s}^{-2}\times 18m$

$v^2=64+36$

$v^2=100$

$v=10m{s}^{-1}$

Thus,

The car is travelling at a speed of 10m/s.