Jack is driving with a pail of water along a straight pathway at a steady 25 m/s when he passes Jill who is parked in her minivan waiting for him. When Jack is beside Jill, she begins accelerating at the rate of 4.0 × 10−3 m/s2 in the same direction that Jack is driving. How long does it take Jill to catch up to Jack?

Question

Jack is driving with a pail of water along a straight pathway at a steady 25 m/s when he passes Jill who is parked in her minivan waiting for him. When Jack is beside Jill, she begins accelerating at the rate of 4.0 × 10−3 m/s2 in the same direction that Jack is driving. How long does it take Jill to catch up to Jack?

Answer 1

d1 = d2

0.5a*t^2 = 25t
0.5a*t = 25
0.5*0.004t = 25
t = 12,500 s. = 3.5 h.

Answer 2

@Henry og bOI

Answer 3

To find out how long it takes for Jill to catch up to Jack, we need to determine the distance between them when Jill starts accelerating and then calculate the time it takes for Jill to cover that distance.

Let's assume the initial distance between Jack and Jill is d. When Jill starts accelerating, Jack will already be some distance away, given by d1 = 25 m/s * t, where t is the time it took for Jill to start accelerating.

Now, let's find out Jill's position as a function of time. We can use the following kinematic equation to calculate the distance Jill covers during her acceleration period:

d2 = ut + (1/2) * acceleration * t^2

Where d2 is the distance covered by Jill during her acceleration, u is her initial velocity (0 m/s since she was parked initially), acceleration is the rate at which she accelerates (4.0 × 10^(-3) m/s^2), and t is the time taken during her acceleration.

Then, Jill's overall distance d can be expressed as the sum of d1 and d2:

d = d1 + d2

Since we want Jill to catch up to Jack, we can set d equal to 0, which gives us:

0 = 25 m/s * t + (1/2) * (4.0 × 10^(-3) m/s^2) * t^2

Simplifying the equation, we get:

0 = 25t + 2.0 × 10^(-3) t^2

Rearranging the equation, we have:

2.0 × 10^(-3) t^2 + 25t = 0

Now, we can solve this quadratic equation for t using the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / 2a

In our case, a = 2.0 × 10^(-3), b = 25, and c = 0. Plugging these values into the quadratic formula, we have:

t = (-25 ± √(25^2 - 4 * 2.0 × 10^(-3) * 0)) / (2 * 2.0 × 10^(-3))

Simplifying further:

t = (-25 ± √(625)) / (4.0 × 10^(-3))

t = (-25 ± 25) / (4.0 × 10^(-3))

t = 0 or t = -6250

Since time cannot be negative, the only value of t that makes sense in this context is t = 0.

Therefore, it will take Jill no time to catch up to Jack.

Answer 4

To find out how long it will take for Jill to catch up to Jack, we first need to determine the distance covered by both Jack and Jill.

Let's define some variables:
- t: Time taken by Jill to catch up to Jack
- d: Distance covered by Jack when Jill catches up to him (also the distance covered by Jill)

We know that Jack's speed is 25 m/s and that Jack and Jill start at the same position. So, after t seconds, Jack will have covered a distance of 25t meters.

To determine Jill's distance at time t, we need to consider her initial velocity, which is zero. Her acceleration is given as 4.0 × 10^(-3) m/s^2.

Using the equations of motion:

Jill's displacement = Initial Velocity * Time + (1/2) * Acceleration * Time^2

Since Jill's initial velocity is 0, the equation simplifies to:

Jill's displacement = (1/2) * Acceleration * Time^2

Plugging in the values:

d = (1/2) * (4.0 × 10^(-3)) * (t)^2
d = 2 * 10^(-3) * t^2

Now, we equate the distances covered by Jack and Jill:

25t = 2 * 10^(-3) * t^2

Simplifying the equation:

25t = 2 * 10^(-3) * t^2
25 = 2 * 10^(-3) * t
t = 25 / (2 * 10^(-3))
t = 25 * 10^3 / 2

So, it will take Jill approximately 12,500 seconds to catch up to Jack.