When starting from rest at the bottom of a straight road with constant upward slope, Joan bicycles to the top 50.0 s ahead of Sally, whose travel time is 5.00 minutes. What is the ratio of Joan's acceleration to Sally's acceleration?
A) 0.694
B) 0.833
C) 1.20
D) 1.44
E) 6.00

Question

Quizplus · Accepted Answer

Let's call the distance to the top of the slope "d". 
Sally's average speed is: 
v = d / t = d / 300 s

Joan's average speed is: 
v = d / (t - 50s)

Since we know that Joan started from rest and we want to find the ratio of her acceleration to Sally's acceleration, it's useful to use the kinematic equation: 
d = 1/2 * a * t^2

For Sally: 
d = 1/2 * a_sally * (300 s)^2 
a_sally = 2d / (300 s)^2

For Joan: 
d = 1/2 * a_joan * (t-50s)^2 
2d / (t-50s)^2 = a_joan 
Notice that we can substitute (d / (t-50s)) for Joan's average speed, which gives: 
2 * (d / (t-50s)) / (t-50s) = a_joan

Now we can find the ratio of their accelerations: 
a_joan / a_sally = [2 * (d / (t-50s)) / (t-50s)] / [2d / (300 s)^2] 
a_joan / a_sally = [300^2 / (t-50s)^2] * [(d / (t-50s)) / d] 
a_joan / a_sally = [300^2 / (t-50s)^2] * (1 / (t-50s))

Plugging in the values: 
a_joan / a_sally = [300^2 / (250s)^2] * (1 / 250s) 
a_joan / a_sally = 1.44

When Starting from Rest at the Bottom of a Straight