Q24PE - College Physics (Urone)

Question

While entering a freeway, a car accelerates from rest at a rate of \({\bf{2}}{\bf{.40}}\;{\bf{m/}}{{\bf{s}}^{\bf{2}}}\)  for 12.0 s. (a) Draw a sketch of the situation. (b) List the knowns in this problem. (c) How far does the car travel in those 12.0 s? To solve this part, first identify the unknown, and then discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, check your units, and discuss whether the answer is reasonable. (d) What is the car’s final velocity? Solve for this unknown in the same manner as in part (c), showing all steps explicitly.

Step-by-Step Solution

Answer

(a) The sketch of the situation is given in figure (1).

(b) The known variables for problem are initial velocity, acceleration of car and duration for acceleration.

(c) The distance travelled by the car is \({\bf{172}}{\bf{.8}}\;{\bf{m}}\).

(d) The final velocity of the car is \({\bf{28}}{\bf{.8}}\;{\bf{m/s}}\).

Step 1: Sketch for the situation

(a)

Given Data:

The initial speed of commuter train is \(u = 0\) 

The acceleration of car is \(f = 2.40\;{\rm{m}}/{{\rm{s}}^2}\) 

The time for acceleration is \(T = 12\;{\rm{s}}\) 

The time for acceleration and deceleration of commuter train is found by using first equation of motion.

The sketch for the situation is given in below figure:

Figure (1)

Therefore, the duration for acceleration of commuter train is \(16.5\;{\rm{s}}\).

Step 2: Determination of deceleration of rocket sled (b)

The known variables for car’s initial position are initial velocity \(u = 0\), time for the acceleration of car\(T = 12\;{\rm{s}}\) and acceleration of car \(f = 2.40\;{\rm{m}}/{{\rm{s}}^2}\).

Step 3: Determination of distance covered by car (c)

The distance covered by car is given as:

\(d = ut + \frac{1}{2}f{T^2}\) 

Here, \(d\) is the distance covered by car.

Substitute all the values in the above equation.

\(\begin{array}{l}d = \left( 0 \right)\left( {12\;{\rm{s}}} \right) + \frac{1}{2}\left( {2.40\;{\rm{m}}/{{\rm{s}}^2}} \right){\left( {12\;{\rm{s}}} \right)^2}\\d = 172.8\;{\rm{m}}\end{array}\) 

Therefore, the distance covered by car is \(172.8\;{\rm{m}}\).

Step 4: Determination of distance covered by car (d)

The final velocity of car is given by:

\({v^2} = {u^2} + 2ad\) 

Substitute all the values in the above equation.

\(\begin{array}{c}{v^2} = {\left( 0 \right)^2} + 2\left( {2.40\;{\rm{m}}/{{\rm{s}}^2}} \right)\left( {172.8\;{\rm{m}}} \right)\\v = 28.8\;{\rm{m}}/{\rm{s}}\end{array}\) 

Therefore, the final velocity of car is \(28.8\;{\rm{m}}/{\rm{s}}\).