Question

An object experiences an acceleration of -6.8 m/s². As a result, it accelerates from 54 m/s to a complete stop. How much distance did it travel during that acceleration? ​

Answer 1

Given :-

• Initial velocity (u) = 54 m/s
• Final velocity (v) = 0 m/s
• Acceleration (a) = -6.8 m/s²

Negative signs means retardation

To find :-

• Distance travelled (s)

According to the question,

By using Newtons third equation of motion,

→ v² = u² + 2as

Where,

• v = Final velocity
• u = Initial velocity
• s = Distance
• a = Acceleration

→ Substituting the values,

→ (0)² = (54)² + 2 × (-6.8) × s

→ 0 = 2916 + (-13.6s)

→ 0 - 2916 = - 13.6s

→ -2916 = - 13.6s

→ 2916 ÷ 13.6 = s

→ 214.41 = s

.°. Distance travelled is 214.41 metre.

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Answer 2

Given :-

Acceleration of the object, a = -6.8m/s²

Initial Velocity of the object, u = 54 m/s

Final Velocity of the object, v = 0 m/s

To find :-

The distance travelled by the object

Solution :-

As we are provided with constant accerlation (a) initial Velocity (u) and final velocity (v) we can use 3rd equation of motion .

so, using 3rd equation of motion .i.e.,

➠ v² - u² = 2as

here,

v denotes final velocity,u denotes initial Velocity,a denotes accerlation and s denotes distance covered

so, substituting all the given values .i.e.,

➠ (0)² - (54)² = (2)(-6.8)(s)

➠ -2916 = -13.6s

➠ s = 2916/13.6

➠ s = 214.4 m

thus, the distance travelled by the object is 214.4m.

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