Formula one racers speed up much more quickly than normal passenger vehicles, and they also can stop in a much shorter distance. a formula one racer traveling at 90 m/s can stop in a distance of 110 m.what is the magnitude of the car's acceleration as it slows during braking?

Respuesta :

Answer: [tex]-36.81ms^{-2}[/tex]

Using equation of motion:

[tex]v^2-u^2=2as[/tex]

Where, v is the final velocity, u is the initial velocity, a is the acceleration and s is the displacement.

The initial velocity is given as  90 m/s

The final velocity is = 0 m/s

It covers a distance of 110 m before stopping.

[tex]\Rightarrow a=\frac{v^2-u^2}{2s}\\ \Rightarrow a=\frac{0^2-90^2}{2\times110}=-36.81ms^{-2}[/tex]

In order to slow down, the car would decelerate at [tex]-36.81ms^{-2}[/tex].



The acceleration of the car after the brakes are applied is [tex]\boxed{-36.82\text{ m/s}^2}[/tex]. The negative sign of the acceleration shows that it is the deceleration of the car.

Explanation:

The initial velocity of the car is [tex]90\text{ m/s}[/tex].

Finally the car comes to rest. So, the final velocity of the car is [tex]0\text{ m/s}[/tex].

The distance covered by the car during deceleration is [tex]110\text{ m}[/tex].

Concept:

The acceleration of the car is defined as the rate of change of velocity of the car. The change in velocity of the car every second is defined as the acceleration of the body.

As the brakes are applied, the motion of the car under deceleration is given by the third equation of motion.

The mathematical expression for the third equation of motion is given as:

[tex]\boxed{v_f^2=v_i^2+2aS}[/tex]

Here, [tex]v_i[/tex] is the initial velocity of the car, [tex]v_f[/tex] is the final velocity of the car, [tex]a[/tex] is the acceleration of the car and [tex]S[/tex] is the distance covered.

Substitute the values in the above expression.

[tex]\begin{aligned}(0)^2&=(90)^2+2.a.(110)\\a&=\dfrac{0-8100}{2\times110}\\&=\dfrac{-8100}{220}\text{ m/s}^2\\&=-36.82\text{ m/s}^2\end{aligned}[/tex]

Thus, the acceleration of the car after the brakes are applied is [tex]\boxed{-36.82\text{ m/s}^2}[/tex]. The negative sign of the acceleration shows that it is the deceleration of the car.

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Answer Details:

Grade: High School

Subject: Physics

Chapter: Force and acceleration

Keywords:

Car, acceleration, brakes, speed up, racer, vehicle, shorter distance, magnitude, initial velocity, final, slows, travelling, 90 m/s.