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

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,

$\begin{gathered} t=37\text{ min} \\ t=\frac{37}{60} \\ t=0.62\text{ h} \end{gathered}$

Thus, the average acceleration of the robber is,

$a=\frac{v-u}{t}$

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

Substituting the known values,

$\begin{gathered} a=\frac{101.0-133.0}{0.62} \\ a=\frac{-32}{0.62} \\ a=-51.61mi/h^2 \end{gathered}$

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.