Question

At 3:00 P.M., a bank robber is spotted driving north on I-15 at milepost 124. His speed is 133.0 mi/h. At 3:37 P.M., he is spotted at milepost 181 doing 101.0 mi/h. (Assume a straight highway). Assume north to be positive.the bank robber’s displacement during this time interval = 57mithe bank robber’s average velocity during this time interval = 92.432 mi/hWhat is his average acceleration during this time interval? Enter a positive value if the direction is toward north and enter a negative value if the direction is toward south.

Answer 1

The bank robber's average acceleration during the time interval of 3:00 P.M. to 3:37 PM is -119.355 mi/h², indicating a southward direction.

Step-by-step explanation:

To calculate average acceleration, we can use the formula:

average acceleration = (final velocity - initial velocity) / time

Given that the bank robber's average velocity during this time interval is 92.432 mi/h and the time interval is 37 minutes or 0.62 hours, we can calculate the final velocity by multiplying the average velocity by the time interval:

final velocity = average velocity x time interval

final velocity = 92.432 mi/h x 0.62 h = 57.257 mi

Now, we can substitute the values into the average acceleration formula:

average acceleration = (57 mi/h - 133 mi/h) / 0.62 h = -119.355 mi/h²

Since the direction is toward south, the average acceleration during this time interval is -119.355 mi/h².

Answer 2

Given data:

The initial velocity of the robber towards the north is 133.0 mi/h.

The final velocity of the robber towards the north is 101.0 mi/h.

The starting time of the robber is 3:00 PM.

The final time of the robber is 3:37 PM.

Solution:

The time taken by the robber in the given ride is,

`\begin{gathered} t=37\text{ min} \\ t=(37)/(60) \\ t=0.62\text{ h} \end{gathered}`

Thus, the average acceleration of the robber is,

`a=(v-u)/(t)`

where v is the final velocity and u is the initial velocity,

Substituting the known values,

`\begin{gathered} a=(101.0-133.0)/(0.62) \\ a=(-32)/(0.62) \\ a=-51.61mi/h^2 \end{gathered}`

Here, the negative sign indicates the direction of acceleration is towards the south.

Hence, the average acceleration of the robber is -51.61 miles per hour squared.