A car’s brakes decelerate it at a rate of -2.40 m/s2. If the car is originally travelling at 13 m/s and comes to a stop, then how far, in meters, will the car travel during that time?

Respuesta :

Answer:

Approximately [tex]35.2\; \rm m[/tex].

Explanation:

Given:

Initial velocity: [tex]u = 13\; \rm m \cdot s^{-1}[/tex].

Acceleration: [tex]a = -2.40\; \rm m \cdot s^{-2}[/tex] (negative because the car is slowing down.)

Implied:

Final velocity: [tex]v = 0\; \rm m \cdot s^{-1}[/tex] (because the car would come to a stop.)

Required:

Displacement, [tex]x[/tex].

Not required:

Time taken, [tex]t[/tex].

Because the time taken for this car to come to a full stop is not required, apply the SUVAT equation that does not involve time:

[tex]\begin{aligned} x &= \frac{v^2 - u^2}{2\, a} \\ &= \frac{{\left(0\; \rm m \cdot s^{-1}\right)}^2 - {\left(13\; \rm m \cdot s^{-1}\right)}^2}{2\times \left(-2.40\; \rm m\cdot s^{-2}\right)} \approx 35.2\; \rm m \end{aligned}[/tex].

In other words, this car would travel approximately [tex]35.2\; \rm m[/tex] before coming to a stop.