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Chapter 3: Problem 39

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

Short Answer

Expert verified
Answer: To find the distance travelled by the body, we can use the derived formula \(d = s \times t\) and substitute the given values of speed (\(s = 10\) m/s) and time (\(t = 5\) s): \(d = 10 \times 5\) \(d = 50\) meters The body travels a distance of 50 meters.

Step by step solution

01

Identify the given formula

The given formula to calculate speed is: Speed \(= \frac{\text{Distance travelled}}{\text{Time taken}}\)
02

Write down the given values

We are given: - speed of the body (to be represented as \(s\)) - time taken (to be represented as \(t\)) And we need to find the distance travelled (to be represented as \(d\)).
03

Rearrange the formula to find the distance travelled

The original formula is: \(s = \frac{d}{t}\) To find the distance travelled \(d\), we can simply multiply both sides of the equation by \(t\): \(d = s\times t\)
04

Use the derived formula to find the distance travelled

With the formula \(d = s \times t\), we can now determine the distance travelled by the body, given the speed of the body (\(s\)) and the time taken (\(t\)). Simply substitute the given values of speed and time into the formula to calculate the distance travelled.

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Most popular questions from this chapter

(i) The unit of speed and velocity is \(\mathrm{m} \mathrm{s}^{-1}\). (ii) If a body is moving in a straight line path then its speed is equal to velocity \(50 \mathrm{~m} \mathrm{~s}^{-1}=50 \times \frac{18}{5} \mathrm{~km} \mathrm{~h}^{-1}=180 \mathrm{~km} \mathrm{~h}^{-1}\)

Fill in the Blanks. velocity When a car moves with constant velocity, its average velocity is equal to its velocity.

The speed and average speed of the vehicle can be equal if the vehicle moves with uniform speed or constant speed.

Fill in the Blanks. length \(\mathrm{T} \alpha \sqrt{\ell}\) Time period of a simple pendulum depends on its length.

From the given figure, by pythogorous theorem \(\mathrm{PR}^{2}=\mathrm{PQ}^{2}+\mathrm{R} \mathrm{Q}^{2}\) \(\mathrm{PR}^{2}=4^{2}+3^{2}\) \(\mathrm{PR}^{2}=16+9\) \(\mathrm{PR}=\sqrt{25}\) \(\mathrm{PR}=5 \mathrm{~m}\) Average speed \(=\frac{\text { total distance }}{\text { total time }}=\frac{4 \mathrm{~m}+3 \mathrm{~m}+5 \mathrm{~m}}{20 \mathrm{~s}+10 \mathrm{~s}+30 \mathrm{~s}}=\frac{12 \mathrm{~m}}{60 \mathrm{~s}}=0.2 \mathrm{~m} \mathrm{~s}^{-1}\)
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