Q25PE - College Physics (Urone)

Question

At the end of a race, a runner decelerates from a velocity of 9.00 m/s at a rate of \({\bf{2}}\;{\bf{m/}}{{\bf{s}}^{\bf{2}}}\). (a) How far does she travel in the next 5.00 s? (b) What is her final velocity? (c) Evaluate the result. Does it make sense?

Step-by-Step Solution

Answer

(a) The distance travelled by racer is \({\bf{20}}\;{\bf{m}}\).

(b) The final velocity of racer is \({\bf{1}}\;{\bf{m/s}}\).

(c) The result does not make any sense.

Step 1: Determination of distance travelled by racer

(a)

Given Data:

The initial velocity of racer is \(u = 9\;{\rm{m}}/{\rm{s}}\) 

The deceleration of racer is \(a = 2\;{\rm{m}}/{{\rm{s}}^2}\) 

The time for distance travelled is \(t = 5\;{\rm{s}}\) 

The distance travelled by the racer can be found by using second equation of motion with deceleration.

The distance travelled by the racer is given as:

\(d = ut - \frac{1}{2}a{t^2}\) 

Here, \(a\) is the deceleration of racer.

Substitute all the values in the above equation.

\(\begin{array}{l}d = \left( {9\;{\rm{m}}/{\rm{s}}} \right)\left( {5\;{\rm{s}}} \right) - \frac{1}{2}\left( {2\;{\rm{m}}/{{\rm{s}}^2}} \right){\left( {5\;{\rm{s}}} \right)^2}\\d = 20\;{\rm{m}}\end{array}\) 

Therefore, the distance travelled by racer is \(20\;{\rm{m}}\).

Step 2: Determination of final velocity of racer

(b)

The final velocity of racer is given as:

\({v^2} = {u^2} - 2ad\) 

Here, \(v\) is the final velocity of racer.

Substitute all the values in the above equation.

\(\begin{array}{c}{v^2} = {\left( {9\;{\rm{m}}/{\rm{s}}} \right)^2} - 2\left( {2\;{\rm{m}}/{{\rm{s}}^2}} \right)\left( {20\;{\rm{m}}} \right)\\v = 1\;{\rm{m}}/{\rm{s}}\end{array}\) 

Therefore, the final velocity of racer is \(1\;{\rm{m}}/{\rm{s}}\).

Step 3: Evaluation of result

(c)

The time for complete stop of racer with deceleration is given as:

\(V = u - aT\) 

Here, \(V\) is the velocity of racer at stop and its value is zero, \(T\) is the duration necessary for stop.

Substitute all the values in the above equation.

\(\begin{array}{l}0 = 9\;{\rm{m}}/{\rm{s}} - \left( {2\;{\rm{m}}/{{\rm{s}}^2}} \right)T\\T = 4.5\;{\rm{s}}\end{array}\) 

The duration for complete stop of racer is less than duration for distance travelled by racer so this contradiction in time does not make any sense for the motion of racer.