Question

A particle moving in a straight line covers half the distance with speed of 3 m/s.The other half of the distance is covered in two equal time intervals with speed 4.5m/s amd 7.5m/s respectively. The average speed of the particle during this motion is​

Answer 1

Answer:

it's 4m/s

Explanation:

time=s/v

t1=s/2/3

t1=s/6

t2=t3

t2+t3=s/12

total time=s/6+s/12

=s/s/4

=4m/s

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Answer 2

Answer:

$\boxed{\mathfrak{Average \ speed \ (v_{avg} )= 4 \ m/s }}$

Explanation:

Let the total distance covered by particle be 'd'

Velocity to cover first half distance is given as 3 m/s

Let time taken to cover first half distance be $\rm t_1$

$\rm \implies t_1 = \dfrac{d}{2 \times 3} \\ \\ \rm \implies t_1 = \dfrac{d}{6}$

Velocity to cover second half distance is given as 4.5 m/s & 7.5 m/s.

Time interval for 4.5 m/s & 7.5 m/s are equal let it be 't'

Let the distance travelled in 4.5 m/s be $\rm d_2$ & 7.5 be $\rm d_3$

So,

$\rm \implies t = \dfrac{ d_2}{4.5} \: \: \: \: \: ...eq_1 \\ \\ \rm \implies t = \dfrac{d_3}{7.5} \: \: \: \: \: ...eq_2$

$\rm d_2$ & $\rm d_3$ is equal to second half distance i.e.

$\rm d_2 + d_3 = \dfrac{d}{2} \: \: \: \: \: ...eq_3$

From $\rm eq_1$ & $\rm eq_2$ we can find $\rm d_2$ & $\rm d_3$ and substitute it in $\rm eq_3$

$\rm \implies 4.5t + 7.5t = \frac{d}{2} \\ \\ \rm \implies 12t = \frac{d}{2} \\ \\ \rm \implies t= \dfrac{d}{2 \times 12} \\ \\ \rm \implies t= \dfrac{d}{24}$

Total time = $\rm t_1 + t + t$

$\bf Average \: speed \: ( v_{avg})= \dfrac{Total \: distance \: travelled}{Total \: time \ taken}$

$\rm \implies v_{avg} = \frac{d}{t_1 + t + t} \\ \\ \rm \implies v_{avg} = \dfrac{d}{ \dfrac{d}{6} + \dfrac{d}{24 } + \dfrac{d}{24 } } \\ \\ \rm \implies v_{avg} = \dfrac{d}{ \dfrac{4d + d + d}{24} } \\ \\ \rm \implies v_{avg} = \dfrac{\cancel{d}}{ \dfrac{6 \cancel{d}}{24} } \\ \\ \rm \implies v_{avg} = \dfrac{24}{6} \\ \\ \rm \implies v_{avg} = 4 \: m {s}^{ - 1}$