Q76P You are the design engineer in c... [FREE SOLUTION]

Chapter 7: Q76P (page 170)

You are the design engineer in charge of the crashworthiness of new automobile models. Cars are tested by smashing them into fixed, massive barriers at 45 km/h. A new model of mass 1500 kg takes 0.15 s from the time of impact until it is brought to rest. (a) Calculate the average force exerted on the car by the barrier. (b) Calculate the average deceleration of the car in g鈥檚.

Short Answer

Expert verified

The results for parts (a) and (b) are \( - 1.25 \times {10^5}\;{\rm{N}}\) and \(8.49g\), respectively.

Step by step solution

Understanding the average force

Use the relationship between the change in momentum of a particular particle and elapsed time to calculate the average force applied to the car.

Given data

Given data:

The mass of the car is\(m = 1500\;{\rm{kg}}\).

The speed of the car is\(v = 45\;{\rm{km/h}}\).

The time is \(t = 0.15\;{\rm{s}}\).

Find the average force applied by the barrier

The relation of average force can be written as follows:

\(\begin{array}{l}F = \frac{{m\Delta v}}{t}\\F = \frac{{m\left( {{v_0} - v} \right)}}{t}\end{array}\)

Here,\(\Delta v\)is the change in speed of the car, and\({v_0}\)is the initial speed whose value is zero.

Plugging the values in the above equation,

\(\begin{array}{l}F = \left[ {\frac{{1500\;{\rm{kg}}\left( {0 - 45\;{\rm{km/h}} \times \frac{{1\;{\rm{m/s}}}}{{3.6\;{\rm{km/h}}}}} \right)}}{{0.15\;{\rm{s}}}}} \right]\\F = - 1.25 \times {10^5}\;{\rm{N}}\end{array}\).

Thus, \(F = - 1.25 \times {10^5}\;{\rm{N}}\) is the required average force.

Calculate the deceleration of the car

The relation to calculate deceleration can be written as follows:

\(F = ma\)

Plugging the values in the above equation,

\(\begin{array}{c} - 1.25 \times {10^5}\;{\rm{N}} = \left( {1500\;{\rm{kg}}} \right)a\\a = \left( {83.3\;{\rm{m/}}{{\rm{s}}^2} \times \frac{{1\;{\rm{g}}}}{{9.8\;{\rm{m/}}{{\rm{s}}^2}}}} \right)\\a = 8.49g\end{array}\)

Thus, \(a = 8.49g\) is the deceleration.