Question

a student jobs on a track of 150 m long the student starts jogging on the track and reaches the end of the track in one minute 30 second and come back to the middle off the track in 1 minute what is the average velocity of the student
a)0m/sec
b)0.5m/sec
c)1.25m/sec
d)1.5m/sec​

Answer 1

Answer:

Track = 150 m

Finding the average speed :

Case 1 :

Student jogs from starting point to the ending point.

Distance travelled = 150 m

Time taken = 1 m 30 s = 60 s + 30 s = 90 s

Now, Speed = 150/90 = 15/9 = 1.6 m/s

So, the speed in case 1 = 1.6 m/s

Case 2 :

Student jogs from ending point to the middle of the track.

Distance travelled = 150/2 = 75 m

Time taken = 1 m = 60 s

Now, Speed = 75/60 = 1.25 m/s

So, the speed in case 2 = 1.25 m/s

Now, Average speed of the student :

⇒ (Initial Speed+Final Speed)/2

⇒ (1.6+1.25)/2

⇒ 2.85/2

⇒ 1.42 m/s

Finding the average velocity :

Displacement = 150 - 75 = 75 m

Time taken = 90s + 60s = 150 s

Average Velocity = ∆x/t

⇒ 75/150 m/s

⇒ 0.5 m/s

Hence, the average velocity of the student b) 0.5m/sec.

Answer 2

Answer:

Given :-

• A student jobs on a track of 150 m long the student starts jogging on the track and reaches the end of the track in one minute 30 second and come back to the middle of the track in 1 minutes.

To Find :-

• What is the average velocity of the student.

Solution :-

${\normalsize{\bold{\purple{\underline{\bigstar\: In\: the\: 1^{st}\: case\: :-}}}}}$

Given :

$\leadsto$ Distance Covered = 150 m

$\leadsto$ Time Taken

$\implies \sf 1\: minutes\: 30\: seconds$

$\implies \sf 60\: seconds + 30\: seconds\: \: \bigg\lgroup \sf\bold{1\: minutes =\: 60\: seconds}\bigg\rgroup$

$\implies \sf \bold{90\: seconds}$

$\leadsto$ Speed

As we know that,

$\clubsuit$ Speed Formula :

$\mapsto \sf\boxed{\bold{\pink{Speed =\: \dfrac{Distance\: Covered}{Time\: Taken}}}}$

Then,

$\implies \sf \dfrac{15\cancel{0}}{9\cancel{0}}$

$\implies \sf \dfrac{15}{9}$

$\implies \sf\bold{1.6\: m/s}$

${\normalsize{\bold{\purple{\underline{\bigstar\: In\: the\: 2^{nd}\: case\: :-}}}}}$

Given :

$\leadsto$ Distance Covered

$\implies \sf \dfrac{\cancel{150}}{\cancel{2}}$

$\implies \sf\bold{75\: m}$

$\leadsto$ Time Taken :

$\implies \sf 1\: minutes$

$\implies \sf 60\: seconds\: \: \bigg\lgroup \sf\bold{1\: minutes =\: 60\: seconds}\bigg\rgroup$

$\implies \sf\bold{60\: seconds}$

$\leadsto$ Speed :

$\implies \sf \dfrac{75}{60}$

$\implies \sf\bold{1.25\: m/s}$

Now, we have to find the average speed :

$\mapsto \sf\boxed{\bold{\pink{Average\: Speed =\: \dfrac{Total\: Speed}{2}}}}$

$\implies \sf Average\: Speed =\: \dfrac{1.6 + 1.25}{2}$

$\implies \sf Average\: Speed =\: \dfrac{2.85}{2}$

$\implies \sf\bold{\green{Average\: Speed =\: 1.42\: m/s}}$

Now, we have to find the average velocity :

As we know that,

$\mapsto \sf\boxed{\bold{\pink{Average\: Velocity =\: \dfrac{Displacement}{Time\: Taken}}}}$

Given :

$\leadsto$ Displacement :

$\implies \sf 150 - 75$

$\implies \sf\bold{75\: m}$

$\leadsto$ Time Taken :

$\implies \sf 90\: seconds + 60\: seconds$

$\implies \sf \bold{150\: seconds}$

According to the question by using the formula we get,

$\longrightarrow \sf Average\: Velocity =\: \dfrac{75}{150}$

$\longrightarrow \sf\bold{\red{0.5\: m/s}}$

$\therefore$ The average velocity of the student is 0.5 m/s .

Hence, the correct options is option no (b) 0.5 m/s .