A roller coaster starts down a hill at 10 m/s. Three seconds later, its speed is 32 m/s. What is the roller coaster’s acceleration?
a = vf  vi
t
a = vf  vi
t
1 Answer
Aug 3, 2017
Answer:
Explanation:
We're asked to find the (constant) acceleration of the roller coaster, given its final and initial velocities.
To do this, we can use the kinematics equation
#ul(v_x = v_(0x) + a_xt#
where

#v_x# is the final velocity 
#v_(0x)# is the initial velocity 
#a_x# is the acceleration (what we're trying to find) 
#t# is the time interval
We know:

#v_x = 32# #"m/s"# 
#v_(0x) = 10# #"m/s"# 
#a_x = ?# 
#t = 3# #"s"#
Plugging in known values:
#32# #"m/s"# #= 10# #"m/s"# #+ a_x(3color(white)(l)"s")#
#a_x = (32color(white)(l)"m/s"  10color(white)(l)"m/s")/(3color(white)(l)"s") = color(red)(ulbar(stackrel(" ")(" "7.33color(white)(l)"m/s"^2" "))#