Question

Gretchen runs the first 4.0 km of a race at 5.0 m/s. Then a stiff wind comes up, so she runs the last 1.0 km at only 4.0 m/s.
If she runs fhe same course again, what constant speed would let her finish in the same time as in the first race?

Answer 1

Answer:

The velocity is $v = 4.76 \ m/s$

Explanation:

From the question we are told that

   The first distance is   $d_1 = 4.0 \ km = 4000 \ m$

   The  first speed  is  $v_1 = 5.0 \ m/s$

    The  second distance is  $d_2 = 1.0 \ km = 1000 \ m$

    The  second speed  is  $v_2 = 4.0 \ m/s$

Generally the time taken for first distance is  

      $t_1 = \frac{d_1 }{v_1 }$

        $t_1 = \frac{4000}{5}$

       $t_1 = 800 \ s$

The time taken for second  distance is

           $t_1 = \frac{d_2 }{v_2 }$

        $t_1 = \frac{1000}{4}$

       $t_1 = 250 \ s$

The total time is mathematically represented as

     $t = t_1 + t_2$

=>   $t = 800 + 250$

=>    $t = 1050 \ s$

Generally the constant velocity that would let her finish at the same time is mathematically represented as

      $v = \frac{d_1 + d_2}{t }$

=>    $v = \frac{4000 + 1000}{1050 }$

=>    $v = 4.76 \ m/s$

Answer 2

The constant speed that will let her finish in the same time as in the first race is 4.76 m/s

Determination of the time taken for first 4 Km

• Distance = 4 Km = 4 × 1000 = 4000 m

• Speed = 5 m/s

• Time 1 =?

Time 1 = distance / speed

Time 1 = 4000 / 5

Time 1 = 800 s

Determination of the time taken for the last 1 Km

• Distance = 1 Km = 1 × 1000 = 1000 m

• Speed = 4 m/s

• Time 2 =?

Time 2 = distance / speed

Time 2 = 1000 / 4

Time 2 = 250 s

Determination of the constant speed

• Total distance = 4000 + 1000 = 5000 m

• Total time = 800 + 250 = 1050 s

• Constant speed =?

Constant speed = Total distance / total time

Constant speed = 5000 / 1050

Constant speed = 4.76 m/s

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