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 add the steps

Answer 1

Given that,

• Speed  (or) the Initial Velocity of the car = 40 km/h

• Acceleration of the car = (-2.3 m/s²) {∵The brakes of the car are applied and it is acting in the opposite direction}.

• Time taken for the car = 2.7 seconds

• Distance travelled by the car after applying the brakes = ?

• Final Velocity of the car = ?

we know,

$a=\frac{(v-u)}{t}\\(-2.3 m/s^2)=\frac{(v-40 km/h)}{2.7 sec}\\(-2.3 m/s^2)\times(2.7 sec)=v-40 km/h\\\\6.21 m/s=v-40 km/h [Convert\:\:km/h\to m/s\:\:we\:get ]\\6.21 m/s= v-11.1 m/s\\6.21 m/s+11.1 m/s=v\\17.31=v$

Therefore, the Final Velocity of the car is 17.31 m/s

Given that,

• Acceleration = (-2.3 m/s²) (a)

• Initial Velocity of the car = 11.1 m/s (u)

• Final velocity = 17.31 m/s (v)

• Time = 2.7 sec (t)

• Distance travelled by the car = ? (s)

we know,

$s=ut+\frac{1}{2}at^2\\s=11.1\: m/s^2\times2.7 sec+\frac{1}{2}\times(-2.3 m/s)\times(2.7\: sec)^2\\s=29.97 \:m/s^2+\frac{1}{2}\times16.767 \:m/s\\\\s=29.97 \:m/s^2+8.3835\:m/s\\s=38.3535 \:ms= 38.35 \: m\:(Approximately)$

Therefore, the car travelled 38.35 m during the acceleration.

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