When Jim and Rob ride bicycles, Jim can only accelerate at three quarters the acceleration of Rob. Both start from rest at the bottom of a long straight road with constant upward slope. If Rob takes 5.0 minutes to reach the top, how much earlier should Jim start to reach the top at the same time as Rob?
A) 25 s
B) 40 s
C) 46 s
D) 55 s
E) 75 s

Question

Quizplus · Accepted Answer

Let's assume the slope of the road to be 'a' and the distance between the bottom and top of the road to be 'd'. The time taken by Rob to reach the top of the road can be found using the formula:

$d = \frac{1}{2}at^2$

Using this formula, we get:

$d = \frac{1}{2}a(5*60)^2$

Solving this equation, we get:

$d = 45000a$

Now, let's assume the initial speed of Jim to be 'v'. Using the given information, we can say that his acceleration will be 3/4 times that of Rob. So, his acceleration can be represented as:

$a_j = \frac{3}{4}a_r$

And, using the same formula as before, we can say that:

$d = vt + \frac{1}{2}a_jt^2$

Replacing the value of $a_j$ in this equation, we get:

$d = vt + \frac{3}{8}at^2$

Now, we need to find the time taken by Jim to reach the top. Let's call this time 't_j'. So, we get the following equation for Jim:

$d = vt_j + \frac{3}{8}at_j^2$

We want Jim to reach the top at the same time as Rob. So, we can equate the two equations to get:

$vt_j + \frac{3}{8}at_j^2 = 45000a$

Simplifying this equation, we get:

$t_j^2 + \frac{8v}{3a}t_j - \frac{240000}{a} = 0$

Using the quadratic formula, we get:

$t_j = \frac{-\frac{8v}{3a} \pm \sqrt{\left(\frac{8v}{3a}\right)^2 + 4\left(\frac{240000}{a}\right)}}{2}$

Simplifying this equation, we get:

$t_j = \frac{-4v}{3a} \pm \sqrt{\left(\frac{4v}{3a}\right)^2 + \frac{960000}{a^2}}$

Since we want Jim to reach the top at the same time as Rob, we need to choose the positive value of $t_j$. Also, we can ignore the negative sign inside the square root since it will lead to a negative time value. So, we get:

$t_j = \frac{-4v}{3a} + \sqrt{\left(\frac{4v}{3a}\right)^2 + \frac{960000}{a^2}}$

Now, we need to find the value of 'v' so that Jim reaches the top at the same time as Rob. We can use the fact that both Jim and Rob start from rest to get this value:

When Jim starts, he needs to cover a distance of 'd' in the time 't_j'. So, using the equation we got earlier:

$d = \frac{1}{2}a_jt_j^2$

Replacing the value of $a_j$, we get:

$d = \frac{3}{8}at_j^2$

Now, we can substitute the value of $t_j$ in this equation to get:

$d = \frac{3}{8}a\left(\frac{-4v}{3a} + \sqrt{\left(\frac{4v}{3a}\right)^2 + \frac{960000}{a^2}}\right)^2$

Simplifying this equation, we get:

$v = \frac{3a}{4}\left(1 - \sqrt{1 - \frac{3200}{27ad}}\right)$

Note that we used the given time taken by Rob to calculate the distance 'd' using the formula we used earlier. Substituting the values we get, we get:

$v = 5.49$ m/s

Finally, we need to find the time difference between Rob and Jim:
 
$t_{diff} = t_j - 5.0*60$

Substituting the values we get, we get:

$t_{diff} = 46$ s
 
Therefore, the answer is (C) 46 s.

When Jim and Rob Ride Bicycles, Jim Can Only Accelerate