Question

A car traveling at 40.0 km/h accelerates at -2.3 m/s2 for 2.7 s. How far has it traveled in that time? What is its final velocity?Please include steps

Answer 1

Given that,

• Speed of the car = 40 km/h = 40×5/18 m/s = 11.1 m/sec

• Acceleration (or) Retardation due to brakes applied = (-2.3 m/s²)

• Time = 2.7 sec

• Distance travelled by the car after applying brakes = ?

• Final velocity of the car = ?

we know,

$s=ut+\frac{1}{2}at^2\\s = 11.1\times2.7+\frac{1}{2}\times(-2.3)\times(2.7)^2\\s= 29.97+(-3.3835)\\s=29.97-3.3835\\s=26.5865m$

∴ The car travelled 26.5865 km after applying the brakes.

we know,

• Initial Velocity = 11.1 m/s

• Acceleration = (-2.3 m/s²)

• Time = 2.7

• Final Velocity = ?

$a=\frac{v-u}{t}\\(-2.3)=\frac{v-11.1}{2.7}\\v=(-2.3)+11.1\times2.7\\v=(-2.3)+29.97\\v=-2.3+29.97\\v=27.67$

∴ The Final Velocity of the car is 27.67 m/s

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