Question

maths chapter :- speed ,distance and time question number 5,6,7,8 plz help me​

Answer 1

Question :

How long will it take a bus travelling at 48 km/h to cover a distance of 360 m.

Given :

• Speed = 48 km/h

• Distance = 360 m

To find :

Time Taken to cover the given distance.

Solution :

First let us convert the speed from km/h to m/s.i.e

$\bf{:\implies v = 48 km/h}$

$\bf{:\implies v = \bigg(48 \times \dfrac{5}{18}\bigg)m}$

$\bf{:\implies v = 13.33 (approx) m }$

Hence, the speed of the bus is 13.33 m.

Now using the formula for speed and substituting the values in it, we get :

$:\implies \bf{Speed\:(v) = \dfrac{Distance\:(d)}{Time\:(t)}}$

$:\implies \bf{13.33 = \dfrac{360}{t}}$

$:\implies \bf{t = \dfrac{360}{13.33}}$

$:\implies \bf{t = 27(approx.)}$

$\boxed{\therefore \bf{Time\:taken\:(t) = 27\:s} }$

Hence, the time taken is 27 s.

Question :

A car covers 200 km in 4½ hrs and then another 240 km in 3½ hours . Find it's average speed.

Given :

• Distance $s_{1}$ = 200 km

• Distance $s_{2}$ = 200 km

• Time $t_{1}$ = 4½ hrs or 9/2 hours

• Time $t_{2}$ = 3½ hours or 7/2 hours

To find :

The average speed of the car.

Solution :

Using the formula for average speed of an object and substituting the values in it, we get :

$:\implies \bf{Average\:Speed\:(v) = \dfrac{Total\:Distance\:(d)}{Total\:Time\:(t)}}$

$:\implies \bf{v = \dfrac{s_{1} + s_{2}}{t_{1} + t_{2}}}$

$:\implies \bf{v = \dfrac{200 + 240}{\dfrac{9}{2} + \dfrac{7}{2}}}$

$:\implies \bf{v = \dfrac{440}{\dfrac{9 + 7}{2}}}$

$:\implies \bf{v = \dfrac{440}{\dfrac{16}{2}}}$

$:\implies \bf{v = \dfrac{440}{16} \times 2}$

$:\implies \bf{v = \dfrac{440}{16} \times 2}$

$:\implies \bf{v = \dfrac{880}{16}}$

$:\implies \bf{v = 55}$

$\boxed{\therefore \bf{Average\:speed = 55\:km h^{-1}}}$

Hence, the average speed of the car is 55 km/h

Question :

A truck travels at 40 km/h for 3 h and 35 km/h for 2 h. Find it's average speed.

Given :

• Speed $v_{1}$ = 40 km/h
• Speed $v_{2}$ = 35 km/h

• Distance $t_{1}$ = 3 hours

• Distance $t_{2}$ = 2hours

To find :

The average speed of the truck.

Solution :

Let us find the distance covered by the truck in both the journeys .

Formula used :-

$:\implies \bf{Speed\:(v) = \dfrac{Distance\:(d)}{Time\:(t)}}$

Distance covered :

• For the first journey :

$:\implies \bf{40 = \dfrac{d_{1}}{3}}$

$:\implies \bf{40 \times 3 = d_{1}}$

$:\implies \bf{120 = d_{1}}$

$\underline{\therefore \bf{d_{1} = 120\:km}}$

Hence, the Distance traveled in first journey is 120 km.

• For the second journey :

$:\implies \bf{35 = \dfrac{d_{2}}{2}}$

$:\implies \bf{35 \times 2 = d_{2}}$

$:\implies \bf{70 = d_{2}}$

$\underline{\therefore \bf{d_{2} = 70\:km}}$

Hence, the Distance traveled in second journey is 70 km.

Average speed :

$:\implies \bf{v = \dfrac{d_{1} + d_{2}}{t_{1} + t_{2}}}$

$:\implies \bf{v = \dfrac{120 + 70}{3 + 2}}$

$:\implies \bf{v = \dfrac{190}{5}}$

$:\implies \bf{v = 38}$

$\boxed{\therefore \bf{Average\:speed = 38\:km h^{-1}}}$

Hence, the average speed of the car is 38 km/h.

Question :

A man walks 6 km at 12 km/h and returns at 10 km/h. Find his average speed.

Given :

• Distance = 6 km (Here distance will be same in both the cases)

• Speed $v_{1}$ = 12 km/h
• Speed $v_{2}$ = 10 km/h

To find :

The average speed of the whole Journey.

Solution :

Let us find the time taken by the man in both the journeys.

Formula used :-

$:\implies \bf{Speed\:(v) = \dfrac{Distance\:(d)}{Time\:(t)}}$

Time Taken :

• For the first Journey :

$:\implies \bf{12 = \dfrac{6}{t_{1}}}$

$:\implies \bf{t_{1} = \dfrac{6}{12}}$

$:\implies \bf{t_{1} = 0.5}$

$\underline{\therefore \bf{t_{1} = 0.5}}$

Hence, the time taken for the first journey is 0.5 s.

• For the second Journey :

$:\implies \bf{10 = \dfrac{6}{t_{2}}}$

$:\implies \bf{t_{2} = \dfrac{6}{10}}$

$:\implies \bf{t_{2} = 0.6}$

$\underline{\therefore \bf{t_{2} = 0.6}}$

Hence, the time taken for the second journey is 0.6 s.

Average speed :

$:\implies \bf{v = \dfrac{d_{1} + d_{2}}{t_{1} + t_{2}}}$

$:\implies \bf{v = \dfrac{6 + 6}{0.5 + 0.6}}$

$:\implies \bf{v = \dfrac{12}{1.1}}$

$:\implies \bf{v = 10.9(approx)}$

$\boxed{\therefore \bf{Average\:speed = 10.9\:km h^{-1}}}$

Hence, the average speed of the car is 10.9 km/h

Answer 2

Answer:

5.

Given Parameters :

• Speed = 48 km/hr = 48 × 5/18 m/s
• Distance = 360 m

Unknown :

• Time taken = ?

Solution :

$:\implies \sf Speed = \dfrac{Distance}{Time}$

$:\implies \sf Time = \dfrac{Distance}{Speed}$

$:\implies \sf Time = \dfrac{360}{48 \times \dfrac{5}{18} }$

$:\implies \sf Time = \dfrac{360 \times 18}{48 \times 5} \: \: \: \Bigg\lgroup \bf{\because Apply \: the \: Fraction \: rule : \dfrac{a}{\frac{b}{c}} = \frac{a \times c}{b} }\Bigg\rgroup$

$:\implies \sf Time = \dfrac{6480}{240}$

$:\implies \underline{ \boxed{\sf Time = 27 \: seconds }}$

$\therefore\underline{\textsf {Time taken by the bus is \textbf{27 seconds}}}.$

_________________....

6.

$\bigstar \: \underline{\:\large{\sf{For \: First \: \textit {200} \: km :}}}$

$\bullet\:\:\textsf{Distance = \textbf{200 km}}$

$\bullet\:\:\textsf{Time at which the first 200 km distance was covered = 4 \dfrac{1}{2} hr = \textbf{ \dfrac{9}{2} hr}}$

$\bigstar \: \underline{\:\large{\sf{For \: Next \: \textit {240} \: km :}}}$

$\bullet\:\:\textsf{Distance = \textbf{240 km}}$

$\bullet\:\:\textsf{Time at which the next 240 km distance was covered = 3 \dfrac{1}{2} hr = \textbf{ \dfrac{7}{2} hr}}$

☯ $\underline{\boldsymbol{According\: to \:the\: Question\:now :}}$

$\dashrightarrow\:\:\sf Average \: Speed = \dfrac{Total \: Distance}{Total \: Time \: taken}$

$\dashrightarrow\:\:\sf Average \: Speed = \dfrac{200 + 240}{ \dfrac{9}{2} + \dfrac{7}{2} }$

$\dashrightarrow\:\:\sf Average \: Speed = \dfrac{440}{8}$

$\dashrightarrow\:\: \underline{ \boxed{\textsf{ \textbf{Average \: Speed = 55 km/hr}}}}$

$\therefore\underline{\textsf {Average Speed will be \textbf{55 km/hr}}}.$

___________________....

7.

$\bigstar \: \underline{\:\large{\sf{For \: First \: \textit {40} \: km :}}}$

$\bullet\:\:\textsf{Distance = \textbf{40 km}}$

$\bullet\:\:\textsf{Time at which the first 40 km distance was covered = \textbf{3 hr}}$

$\bigstar \: \underline{\:\large{\sf{For \: Next \: \textit {35} \: km :}}}$

$\bullet\:\:\textsf{Distance = \textbf{35 km}}$

$\bullet\:\:\textsf{Time at which the first 35 km distance was covered = \textbf{2 hr}}$

☯ $\underline{\boldsymbol{According\: to \:the\: Question\:now :}}$

$\dashrightarrow\:\:\sf Average \: Speed = \dfrac{Total \: Distance}{Total \: Time \: taken}$

$\dashrightarrow\:\:\sf Average \: Speed = \dfrac{40 + 35}{3 + 2} \\ \\ \\</p><p>$

$\dashrightarrow\:\:\sf Average \: Speed = \dfrac{75}{5} \\ \\ \\</p><p>$

$\dashrightarrow\:\: \underline{ \boxed{\textsf{ \textbf{Average \: Speed = 15 km/hr}}}}$

$\therefore\underline{\textsf {Average Speed will be \textbf{15 km/hr}}}.$

_________________.....

8.

Case (I) :

Distance = 6 km

Speed = 12 km/hr

• As we know that Distance is equals to the Speed into time :]

➝ Distance = Speed × Time

➝ 6 = 12 × Time

➝ Time = 6/12

➝ Time = ½ hours

Case (II) :

Distance = 6 km

Speed = 10 km/hr

➝ Distance = Speed × Time

➝ 6 = 10 × Time

➝ Time = 6/10

➝ Time = ⅗ hours

☯ $\underline{\boldsymbol{According\: to \:the\: Question\:now :}}$

$\dashrightarrow\:\:\sf Average \: Speed = \dfrac{Total \: Distance}{Total \: Time \: taken}$

$\dashrightarrow\:\:\sf Average \: Speed = \dfrac{6 + 6}{ \dfrac{1}{2} + \dfrac{3}{5} }$

$\dashrightarrow\:\:\sf Average \: Speed = \dfrac{12}{ \dfrac{5 + 6}{10 } }$

$\dashrightarrow\:\:\sf Average \: Speed = \dfrac{12}{ \frac{11}{10 } }$

$\dashrightarrow\:\:\sf Average \: Speed = \dfrac{12 \times 10}{ 11} \: \: \: \Bigg\lgroup \bf{\because Apply \: the \: Fraction \: rule : \dfrac{a}{\frac{b}{c}} = \frac{a \times c}{b} }\Bigg\rgroup$

$\dashrightarrow\:\:\sf Average \: Speed = \dfrac{120}{ 11}$

$\dashrightarrow\:\: \underline{ \boxed{\textsf{ \textbf{Average \: Speed = 10.91 km/hr}}}}$

$\therefore\underline{\textsf {Average Speed will be \textbf{10.91 km/hr}}}.$