Question

A car moving along straight line covers 1/5 th of total distance with
speed
V1 and remaining part of the distance with speed U2
The average speed of the car over the entire distance is​

Answer 1

Answer:

5V1U2/4V1+U2

Explanation:

LET THE TOTAL DISTANCE BE x

THEREFORE DISTANCE COVERED WITH SPEED V1=x/5

SPEED=V1

THEREFORE TIME REQUIRED =(x/5)/V1

=x/5V1

REMAINING DISTANCE =x-(x/5)

=4x/5

THEREFORE TIME REQUIRED =(4x/5)/U2

=4x/5U2

THEREFORE TOTAL TIME=(4x/5V2)+(x/5V1)

=(4V1x+xU2)/5V1U2

TAKING x COMMON FROM THE NUMERATOR, WE GET

=x(4V1+U2)/5V1U2

THEREFORE, AVERAGE SPEED =x/x{(4V1+U2)/(5V1U2)}

=x(5V1U2)/x(4V1+U2)

=5V1U2/4V1+U2(as x cancels)

Answer 2

Answer:

Average speed of the car over the entire distance is  $\frac{5U_2V_1}{(U_2 + 4V_1)}$ unit.

Explanation:

• Step-1: Lets assume total distance is 's'.

      So, $\frac{s}{5}$ unit distance is covered with a speed of $V_1$

            Time taken = $\frac{Distance}{Speed}$ = $\frac{s}{5V_1}$

• Step-2: Remaining distance = s - $\frac{s}{5}$ = $\frac{4s}{5}$

                  $\frac{4s}{5}$ unit distance is covered with a speed of $U_2$

•             Time taken = $\frac{Distance}{Speed}$ = $\frac{4s}{5U_2}$
• Step-3: Total distance = s

                     Total time = $\frac{s}{5V_1}$ + $\frac{4s}{5U_2}$  = $\frac{s(U_2 + 4V_1)}{5U_2V_1}$

      Average speed = $\frac{Total distance covered}{total time taken}$ = $\frac{s}{\frac{s(U_2 + 4V_1)}{5U_2V_1}}$ = $\frac{5U_2V_1}{(U_2 + 4V_1)}$ unit

• The SI unit of average speed is m/s. Here the unit of distance and speed are not given, so we can't write a particular unit of average speed.

       Average speed of the car over the entire distance is  $\frac{5U_2V_1}{(U_2 + 4V_1)}$ unit.

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