Question

pls ans both the questions.pls​

Answer 1

GivEn :

• uniform speed of train at which it covers 20 km = 60km/h

• in the next 20 km the uniform speed of train = 80 km/h

To find :

• the Average speed of the train.

Formula used -

$\begin{gathered} \sf \: time \: = \frac{distance}{speed} \\ \\ \sf \: ave. \: speed \: = \frac{total \: distance \: covered}{total \: time \: taken} \end{gathered}$

Solution :

in first case,

$\begin{gathered} \sf \: time \: taken \: to \: cover \: 20km \: at \: 60km {h}^{ - 1} \\ \sf \: = \frac{20}{60} = \frac{1}{3} hrs \\ \\ \end{gathered}$

in second case,

$\begin{gathered} \sf \: time \: taken \: to \: cover \: 20km \: at \: 80km {h}^{ - 1} \\ \sf \: = \frac{20}{80} = \frac{1}{4} \\ \\ \end{gathered}$

therefore total time taken :

$\sf \: \frac{1}{3} + \frac{1}{4} = \frac{7}{12}$

total distance covered :

$\begin{gathered} \sf \: 20 + 20 \sf \rarr 40 \end{gathered}$

therefore Ave. speed :

$\begin{gathered} \sf \: using \: formula \: we \: get \: \\ \\ \sf \: ave \: speed = \frac{40}{ \frac{7}{12} } \\ \\ \sf = \frac{40 \times 12}{7} \\ \\ \sf = \: \frac{480}{7} \\ \\ \sf = 68.54 \: km \: {h}^{ - 1} \end{gathered}$

therefore average speed is 68.54 km/h

2) What type of motion is represented by the following time graphs ?

I) Velocity time graph for uniform velocity

II) Velocity time graph for uniform acceleration

III) Velocity time graph for an object starting from rest

Answer 2

Explanation:

GivEn :

uniform speed of train at which it covers 20 km = 60km/h

in the next 20 km the uniform speed of train = 80 km/h

To find :

the Average speed of the train.

Formula used -

\begin{gathered} \begin{gathered} \sf \: time \: = \frac{distance}{speed} \\ \\ \sf \: ave. \: speed \: = \frac{total \: distance \: covered}{total \: time \: taken} \end{gathered} \end{gathered}

time=

speed

distance

ave.speed=

totaltimetaken

totaldistancecovered

Solution :

in first case,

\begin{gathered} \begin{gathered} \sf \: time \: taken \: to \: cover \: 20km \: at \: 60km {h}^{ - 1} \\ \sf \: = \frac{20}{60} = \frac{1}{3} hrs \\ \\ \end{gathered} \end{gathered}

timetakentocover20kmat60kmh

−1

=

60

20

=

3

1

hrs

in second case,

\begin{gathered} \begin{gathered} \sf \: time \: taken \: to \: cover \: 20km \: at \: 80km {h}^{ - 1} \\ \sf \: = \frac{20}{80} = \frac{1}{4} \\ \\ \end{gathered} \end{gathered}

timetakentocover20kmat80kmh

−1

=

80

20

=

4

1

therefore total time taken :

\sf \: \frac{1}{3} + \frac{1}{4} = \frac{7}{12}

3

1

+

4

1

=

12

7

total distance covered :

\begin{gathered} \sf \: 20 + 20 \sf \rarr 40 \end{gathered}

20+20→40

therefore Ave. speed :

\begin{gathered}\begin{gathered} \sf \: using \: formula \: we \: get \: \\ \\ \sf \: ave \: speed = \frac{40}{ \frac{7}{12} } \\ \\ \sf = \frac{40 \times 12}{7} \\ \\ \sf = \: \frac{480}{7} \\ \\ \sf = 68.54 \: km \: {h}^{ - 1} \end{gathered} \end{gathered}

usingformulaweget

avespeed=

12

7

40

=

7

40×12

=

7

480

=68.54kmh

−1

therefore average speed is 68.54 km/h

2) What type of motion is represented by the following time graphs ?

I) Velocity time graph for uniform velocity

II) Velocity time graph for uniform acceleration

III) Velocity time graph for an object starting from rest