Question

An object travelling along a straight path covers 35 m distance in 4 seconds. In next 6 seconds, it covers 50 m distance. What is the average speed of the object? (1) 3.5 m/s (2) 4.5 m/s (3) 7.5 m/s (4) 8.5 m/s​

Answer 1

Given: An object travelling along a straight path covers 35 m distance in 4 seconds. & In next 6 seconds, it covers 50 m distance.

Need to find: Average speed of the object?

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¤ To calculate average speed, We will divide total distance by total time, It is given by –

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$\: \:\underline{ \boxed{\pmb{\frak{{Average\ speed_{(Object)} = \dfrac{D_1 + D_2}{T_1+T_2} }}}}}$

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• D_1 is 35 m & T_1 is 4 secs, D_2 is 50 m & T_2 is 6 secs.

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$\: \: \: \dag \: \: {\underline{{\frak{Substituting\: values\; in \; formula : }}}}$

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$\begin{gathered}:\implies\sf avg_{(Speed)}= \dfrac{35 + 50}{4 + 6}\\\\\\:\implies\sf avg_{(Speed)} = \dfrac{85}{4 + 6}\\\\\\:\implies\sf avg_{(Speed)} = \cancel{\dfrac{85}{10}} \\\\\\:\implies\underline{\boxed{\pmb{\frak{Avg_{(Speed)} = 8.5 ms^{-1} }}}}\;\bigstar\\\\\end{gathered}$

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$\begin{gathered}\:\;\therefore{\underline{\sf{Average \: speed \: of \: the \: object \: is \:{\sf{\pmb{8.5ms^{-1}}}}.}}}\\\end{gathered}$

Answer 2

Answer:

$\green{ \sf \bigstar \: \:Average\: speed \: \: \bar{v} = 8.5 \: m {s}^{ - 1} }$

Explanation:

Given,

Case-1:

$\sf \: \odot \: distance \: \: \: d_1 = 35 \: m \\ \sf \: \odot \: \:time \: \: t_1 \: = 4 \: s \: \: \: \: \: \: \: \: \: \:$

Case-2:

$\sf \: \odot \: \: distance \: \: d_2 = 50 \: m \\ \sf \: \odot \: \: time \: \: t_2 = 6 \: s \: \: \: \: \: \: \: \: \: \:$

To find :

★ Average\: speed of the object

Solution :

We know that,

$\green{ \boxed{ \sf \: Average\: \: speed \: \: \bar{v} \: = \frac{total \: distance \: (D)}{total \: time \: (T)} }}$

Now,

$\sf \: total \: distance \: \: \: D = d_1 + d_2 \\ = 35 + 50 \\ = 85 \: m \: \: \\ \\ \sf \: total \: time \: \: T = t_1 + t_2 \\ = 4 + 6 \\ = 10 \: s$

$\sf \: Average\: speed \: \: \: \bar{v} = \frac{D}{T} \\ = \frac{85}{10} \\ = 8.5 \: m {s}^{ - 1} \\ \\ \green{ \sf \therefore \:Average\: speed \: \: \bar{v} = 8.5 \: m {s}^{ - 1} }$