Question

A roller coaster, traveling with an initial speed of 15 meters per second, decelerates uniformly at –7.0 m/s2 to a full stop. Approximately how far does the roller coaster travel during its deceleration?

Answer 1

Given: u=15 m/s
v=0m/s
a=-7m/s
Formula:
v²=u²+2as
Soln.:
:.0²=(15)²+2*(-7)*s
:.0=225+(-14)*s
:.-225=-14*s
:.s=-225/-14
≈16.1m
Therefore the displacement was 16.1m

Answer 2

Answer:

Distance, d = 16.07 meters

Explanation:

It is given that,

Initial speed of the roller coaster, u = 15 m/s

Deceleration of the roller coaster, $a=-7\ m/s^2$

Finally, it stops, v = 0

Let the roller coaster travel a distance of d during its deceleration. Using the equation of kinematics to find it. It is given by :

$v^2-u^2=2ad$

$d=\dfrac{v^2-u^2}{2a}$

$d=\dfrac{0-(15\ m/s)^2}{2\times (-7\ m/s^2)}$

d = 16.07 meters

So, the roller coaster travel 16.07 meters during its deceleration. Hence, this is the required solution.

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