Question

3. A car is travelling at a velocity of 25 ms^1 when the engine is then turned off, calculate how far the car

will travel before coming to rest if the deceleration is 1.47 ms^ 2?

Answer 1

Given :-

• Initial velocity (u) = 0 m/s (when the engine is turned off its initial velocity is zero)
• Final velocity (v) = 25 m/s
• Acceleration (a) = 1.47 m/s²

To Find :-

• Distance (s) = ?

Solution :-

Using the third equation of motion we have find the distance :

• 2as = v² - u²

Putting the values according to the equation :

• 2 × (1.47) × s = (25)² - (0)²
• 2.94s = 625 - 0
• 2.94s = 625
• s = 625/2.94
• s = 212.6 m Ans.