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Chapter 6: Q82GP (page 138)

In If a 1300-kg car can accelerate from to in 3.8 s, how long will it take to accelerate from to ? Assume the power stays the same, and neglect frictional losses.

Short Answer

Expert verified

The time taken to accelerate the car from \(55\;{\rm{km/h}}\) to \(95\;{\rm{km/h}}\) is 7.5 s.

Step by step solution

01

Definition of power

Power is directly proportional to the work done on an object. It is defined as the rate of doing work or the work done per unit time. Mathematically,power is represented as\(P = \frac{W}{t}\).

02

Identification of given data

The mass of the car is\(m = 1300\;{\rm{kg}}\).

First case:

The velocities of the car are:

\(\begin{aligned}{v_1} &= \left( {35\;{\rm{km/h}}} \right)\left( {\frac{{1000\;{\rm{m}}}}{{1\;{\rm{km}}}}} \right)\left( {\frac{{1\;{\rm{h}}}}{{3600\;{\rm{s}}}}} \right)\\ &= 9.72\;{\rm{m/s}}\end{aligned}\)

\(\begin{aligned}{v_2} &= \left( {65\;{\rm{km/h}}} \right)\left( {\frac{{1000\;{\rm{m}}}}{{1\;{\rm{km}}}}} \right)\left( {\frac{{1\;{\rm{h}}}}{{3600\;{\rm{s}}}}} \right)\\ &= 18.1\;{\rm{m/s}}\end{aligned}\)

The time taken to accelerate the car is\({t_1} = 3.8\;{\rm{s}}\).

Second case:

The velocities of the car are:

\(\begin{aligned}{v_3} &= \left( {55\;{\rm{km/h}}} \right)\left( {\frac{{1000\;{\rm{m}}}}{{1\;{\rm{km}}}}} \right)\left( {\frac{{1\;{\rm{h}}}}{{3600\;{\rm{s}}}}} \right)\\ &= 15.3\;{\rm{m/s}}\end{aligned}\)

\(\begin{aligned}{v_4} &= \left( {95\;{\rm{km/h}}} \right)\left( {\frac{{1000\;{\rm{m}}}}{{1\;{\rm{km}}}}} \right)\left( {\frac{{1\;{\rm{h}}}}{{3600\;{\rm{s}}}}} \right)\\ &= 26.4\;{\rm{m/s}}\end{aligned}\)

03

Step 3:Determination of the work done

The work done on the car in each case is equal to the change in its kinetic energy and is given by:

\({W_1} = \frac{1}{2}m\left( {v_2^2 - v_1^2} \right)\) and\({W_2} = \frac{1}{2}m\left( {v_4^2 - v_3^2} \right)\)

04

Step 4:Determination of power in the first case

Power is defined as the work done per unit time. In the first case,poweris given by:

\(\begin{aligned}{P_1} &= \frac{{{W_1}}}{{{t_1}}}\\ &= \frac{{\frac{1}{2}m\left( {v_2^2 - v_1^2} \right)}}{{{t_1}}}\\ &= \frac{{\frac{1}{2}\left( {1300\;{\rm{kg}}} \right)\left( {{{\left( {18.1\;{\rm{m/s}}} \right)}^2} - {{\left( {9.72\;{\rm{m/s}}} \right)}^2}} \right)}}{{3.8\;{\rm{s}}}}\\ &= 39940\;{\rm{W}}\end{aligned}\)

05

Step 5:Determinationof the time taken to accelerate the car in the secondcase

You can use the power consumed in the first case to determine the time taken. The power consumed is given as:

\(\begin{aligned}{P_1} &= \frac{{{W_2}}}{{{t_2}}}\\{P_1} &= \frac{{\frac{1}{2}m\left( {v_4^2 - v_3^2} \right)}}{{{t_2}}}\\{t_2} &= \frac{{\frac{1}{2}\left( {1300\;{\rm{kg}}} \right)\left( {{{\left( {26.4\;{\rm{m/s}}} \right)}^2} - {{\left( {15.3\;{\rm{m/s}}} \right)}^2}} \right)}}{{39940\;{\rm{W}}}}\\ &= 7.5\;{\rm{s}}\end{aligned}\)

Thus, the time taken to accelerate the car in the second case is 7.5 s.

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