Question

An oil tanker, moving at a rate of 6.1 m/s, puts its engines in reverse to come to a stop in 14 minutes. What would be the acceleration of that oil tanker while it is coming to a stop?

Answer 1

Given:-

• Initial velocity , u = 6.1m/s

• Final velocity , v = 0m/s

• Time taken ,t = 14 s

To Find:-

• Acceleration of the tanker ,a

Solution:-

Conversion of unit

14min = 14×60s = 840s

As we know that acceleration is defined as the rate of change of velocity at per unit time

• a = v-u/t

Substitute the value we get

→ a = 0-6.1/840

→ a = -6.1/840

→ a = -0.0072m/s²

Here negative sign show deceleration

∴ The acceleration of the oil tanker is 0.0072m/s².

Answer 2

$\rm\boxed\checkmark$ Given :-

• Intial Velocity, u = 6.1m/s.

• Final Velocity, v = 0m/s.

• Time Taken, t = 14 seconds

$\rm\boxed\checkmark$ To Find :-

• Acceleration of tanker

$\rm\boxed\checkmark$ Solution :-

Conversion,

14m = 14 × 60 seconds = 840 seconds.

As we know,

$\implies\sf{a = \dfrac{v - u}{t}}$

Substitute,

$\implies\sf{a \: = \dfrac{0 - 6.1}{840}} \\ \implies\sf{a = \frac{-6.1}{840}} \\ \implies\sf{a = -0.00072m/s}$

Here,

Negative sign = declaration.

༶ Acceleration = -0.0007m/s.