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Chapter 2: Problem 55

A lorry driver sees a old man crossing the road. Suddenly he applies brakes and stops the lorry within \(30 \mathrm{~s}\). If the initial speed of the lorry is \(108 \mathrm{~km} \mathrm{~h}^{-1}\), find its deceleration.

Short Answer

Expert verified
Answer: \(a = -\frac{900}{2s}\), where 'a' is the deceleration and 's' is the stopping distance.

Step by step solution

01

Convert the initial speed to the appropriate units

First, we need to convert the initial speed of the lorry which is given in \(\mathrm{km} \mathrm{~h}^{-1}\) to \(\mathrm{m} \mathrm{~s}^{-1}\): \(v = 108 \mathrm{~km} \mathrm{~h}^{-1} \cdot \frac{1000 \mathrm{~m}}{1 \mathrm{~km}} \cdot \frac{1 \mathrm{~h}}{3600 \mathrm{~s}} = 108 \cdot \frac{5}{18} \mathrm{~m} \mathrm{~s}^{-1}\) \(v = 30 \mathrm{~m} \mathrm{~s}^{-1}\)
02

Apply the third equation of motion to find the deceleration

Knowing that the lorry stops, the final velocity will be \(0 \mathrm{~m} \mathrm{~s}^{-1}\). Using the third equation of motion, which is: \(v^2 = u^2 + 2as\) where: \(v = 0\) (final velocity), \(u = 30 \mathrm{~m} \mathrm{~s}^{-1}\) (initial velocity), \(a\) is the deceleration (negative acceleration), \(s = 0\) (distance travelled).
03

Rearrange the equation and solve for deceleration

Plug in the values for \(v\), \(u\), and \(s\) in the equation and solve for \(a\): \(0 = 30^2 + 2a(0 - s)\) \(0 = 900 - 2as\) \(2as = 900\) \(a = \frac{900}{2s}\) Since the lorry comes to a stop, the initial speed is positive, and there is a negative acceleration (deceleration) acting on it. So, we make 'a' negative: \(a = -\frac{900}{2s}\) Now, the only unknown left in our equation is the stopping distance (\(s\)) of the lorry. But we don't need to find that, as our task is just to find the deceleration. So, the lorry's deceleration can be described by the equation: \(a = -\frac{900}{2s}\)

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