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Chapter 3: Problem 53

Let the distance travelled by car with \(42 \mathrm{~km} \mathrm{~h}^{-1}\) for 20 minutes be \(\mathrm{S}\), and distance travelled with 60 \(\mathrm{km} \mathrm{h}^{-1}\) for 30 minutes be \(\mathrm{S}_{2}\) \(\Rightarrow \mathrm{S}_{1}=42 \mathrm{~km} \mathrm{~h}^{-1} \times \frac{1}{3} \mathrm{~h}=14 \mathrm{~km}\) \(\Rightarrow \mathrm{S}_{2}=60 \mathrm{~km} \mathrm{~h}^{-1} \times \frac{1}{2} \mathrm{~h}=30 \mathrm{~km}\) Total distance \(\mathrm{S}=\mathrm{S}_{1}+\mathrm{S}_{2}=14+30=44 \mathrm{~km}\)

Short Answer

Expert verified
Answer: The total distance traveled by the car is 44 km.

Step by step solution

01

Calculate the first distance (S₁)

To calculate the first distance (S₁), we need to multiply the speed (42 km/h) with the time (20 minutes or 1/3 hours). So, we have: S₁ = 42 km/h × (1/3) h
02

Calculate the second distance (S₂)

To calculate the second distance (S₂), we need to multiply the speed (60 km/h) with the time (30 minutes or 1/2 hours). So, we have: S₂ = 60 km/h × (1/2) h
03

Find the total distance (S)

Now that we've calculated both distances S₁ and S₂, we can find the total distance (S) by adding them together: S = S₁ + S₂ Now, we will perform the calculations and find the total distance.
04

Calculate S₁

Let's calculate the first distance (S₁) based on the formula we've written in Step 1: S₁ = 42 km/h × (1/3) h = 14 km
05

Calculate S₂

Let's calculate the second distance (S₂) based on the formula we've written in Step 2: S₂ = 60 km/h × (1/2) h = 30 km
06

Calculate the total distance (S)

Now we will find the total distance traveled (S) according to the formula in Step 3. Adding the results from Step 4 and Step 5, we get: S = S₁ + S₂ = 14 km + 30 km = 44 km The total distance traveled by the car is 44 km.

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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, \(\mathrm{d}_{\mathrm{A}}=\mathrm{v}_{\mathrm{A}} \times \mathrm{t}=\frac{\mathrm{d}}{20} \times \mathrm{t}\) \(\mathrm{d}_{\mathrm{B}}=\mathrm{v}_{\mathrm{B}} \times \mathrm{t}=\frac{\mathrm{d}}{22} \times \mathrm{t}\) \(\Rightarrow \frac{\mathrm{d}_{\mathrm{A}}}{\mathrm{d}_{\mathrm{B}}}=\frac{\frac{\mathrm{d}}{20}(\mathrm{t})}{\frac{\mathrm{d}}{22}(\mathrm{t})}=\frac{22}{20}=\frac{11}{10}=11: 10\)

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

Take a metre scale and measure the length of the string from the point of suspension to the lower tip of the bob \(\left(\ell_{1}\right)\) (b). Now, place the bob over a meter scale and hold it in position with two wooden blocks or stiff cardboards and measure the diameter (D) of the bob (c). Calculate the radius \(\mathrm{R}\) of the bob by dividing diameter by 2 (a). Then the length of the pendulum \(\ell=\left(\ell_{1}-\mathrm{R}\right)(\mathrm{d})\). Consider the formula \(\mathrm{T}\) \(=2 \pi \sqrt{\frac{\ell}{\mathrm{g}}}\) and find the time period of the simple pendulum by substituting the value of \({ }^{\prime} \ell^{\prime}(\mathrm{e})\).

Distance travelled in the first \(20 \mathrm{~min}=\) speed \(\times\) time \(=60 \times \frac{20}{60}=20 \mathrm{~km}\)

Speed \(=\frac{\text { Distance travelled }}{\text { Time taken }}\) \(\Rightarrow\) distance travelled by the body \(=\) speed of the body \(\times\) time taken.
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