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Chapter 3: Problem 55

The initial velocity, \(\mathrm{u}=108 \mathrm{~km} \mathrm{~h}^{-1}=30 \mathrm{~m} \mathrm{~s}^{-1}\) The final velocity, \(\mathrm{v}=0 \mathrm{~m} \mathrm{~s}^{-1}\) The time taken to stop lorry, \(\mathrm{t}=30 \mathrm{~s}\). The acceleration of the lorry, \(\mathrm{a}=\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}}=\frac{0-30}{30}=-1 \mathrm{~m} \mathrm{~s}^{-2}\)

Short Answer

Expert verified
Answer: The acceleration of the lorry is -1 m/s².

Step by step solution

01

Write down the equation of motion

We will use the following equation of motion to calculate the acceleration: a = (v - u) / t
02

Plug in the values

Using the given values, let's insert them into the equation. u = 30 m/s v = 0 m/s t = 30 s a = (0 - 30) / 30
03

Calculate the acceleration

Now, we can perform the calculation to obtain the acceleration of the lorry: a = (-30) / 30
04

Simplify and write the result

By simplifying the equation, we get: a = -1 m/s² So the acceleration of the lorry is -1 m/s².

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Most popular questions from this chapter

(a) Let the distance travelled be 'd'. The speed of \(^{\prime} A^{\prime}, v_{A}=\frac{d}{20} m s^{-1}\) The speed of ' \(\mathrm{B}^{\prime}, \mathrm{v}_{\mathrm{B}}=\frac{\mathrm{d}}{22} \mathrm{~m} \mathrm{~s}^{-1}\). \(\Rightarrow \frac{\mathrm{v}_{\mathrm{A}}}{\mathrm{v}_{\mathrm{B}}}=\frac{\frac{\mathrm{d}}{20}}{\frac{\mathrm{d}}{22}}=\frac{22}{20}=\frac{11}{10}=11: 10\) (b) Let them run for 't's. Then, \(\mathrm{d}_{\mathrm{A}}=\mathrm{v}_{\mathrm{A}} \times \mathrm{t}=\frac{\mathrm{d}}{20} \times \mathrm{t}\) \(\mathrm{d}_{\mathrm{B}}=\mathrm{v}_{\mathrm{B}} \times \mathrm{t}=\frac{\mathrm{d}}{22} \times \mathrm{t}\) \(\Rightarrow \frac{\mathrm{d}_{\mathrm{A}}}{\mathrm{d}_{\mathrm{B}}}=\frac{\frac{\mathrm{d}}{20}(\mathrm{t})}{\frac{\mathrm{d}}{22}(\mathrm{t})}=\frac{22}{20}=\frac{11}{10}=11: 10\)

$$ \begin{array}{l} \mathrm{V}_{\mathrm{A}}=\frac{\mathrm{d}_{\mathrm{A}}}{\mathrm{t}} ; \mathrm{V}_{\mathrm{B}}=\frac{\mathrm{d}_{\mathrm{B}}}{\mathrm{t}} \\ \mathrm{d}_{\mathrm{A}}=80 \mathrm{~m} \text { and } \mathrm{d}_{\mathrm{B}}=100 \mathrm{~m} \\ \mathrm{~V}_{\mathrm{A}}=\frac{80}{\mathrm{t}} ; \mathrm{V}_{\mathrm{B}}=\frac{100}{\mathrm{t}} \Rightarrow \mathrm{V}_{\mathrm{A}}<\mathrm{V}_{\mathrm{B}} \cdot \mathrm{B}^{\prime} \text { is faster than }^{4} \mathrm{~A}^{\prime} . \end{array} $$

The rest and motion are relative. An object at rest with respect to one observer may not be at rest with respect to another observer. The same can be said about motion. For a person inside a bus, the fellow passengers are at rest but the same passengers are in motion with respect to a person standing on the ground.

Consider a body moving with initial velocity u. Let its velocity change to \(\mathrm{v}\), in time ' \(\mathrm{t}^{\prime}\). Then, the change in velocity is \(=\mathrm{v}-\mathrm{u}\). The change in velocity per unit time \(=\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}}\) By definition, the change in velocity per unit time is acceleration, a. Thus, \(\mathrm{a}=\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}}\) or \(\mathrm{v}-\mathrm{u}=\) at \(\mathrm{v}=\mathrm{u}+\mathrm{at}\)

Let the time period of the simple pendulum on the earth be 'T'. Then,, \(2 \pi \sqrt{\frac{l_{\mathrm{m}}}{\mathrm{g}_{\mathrm{m}}}}\) where, \(\ell_{\mathrm{m}}, \mathrm{g}_{\mathrm{m}}\) are length of the simple pendulum and acceleration due to gravity on the moon.
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