A car traveling at 91.0 km/h approaches the turn off for a restaurant 30.0 m ahead. If the driver slams on the brakes with the a

Question

1 month ago

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n acceleration of -6.40m/s^2what will her stopping distance

1 answer:

kicyunya [3.2K] · Answer 1 · 2026-04-01T16:02:04+03:00

Response: 49.92 m

Clarification:

The following equation is relevant in this context:

$V^{2}=V_{o}^{2} +2 a d$

Where:

$V=0 m/s$ denotes the final speed of the vehicle, once it has come to a halt

$V_{o}=91 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=25.27 m/s$ designates the vehicle's initial speed

$a=-6.4 m/s^{2}$ represents the constant deceleration of the car following the driver engaging the brakes

$d$ signifies the stopping distance

By isolating $d$ :

$d=\frac{-V_{o}^{2}}{2a}$

$d=\frac{-(25.27 m/s)^{2}}{2(-6.4 m/s^{2})}$

$d=41.919 m \approx 41.92 m$