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Lyrx [107]
3 years ago
10

A general contractor drives 5 miles to the hardware store in 10 minutes. He shops for needed tools for 30 minutes, then drives t

o his customer’s home at an average speed of 45 mph. He arrives 20 minutes later. Calculate the average speed from the time the contractor leaves his office to the time he arrives at his customer’s home.
Physics
1 answer:
Alex17521 [72]3 years ago
3 0

Given parameters:

Distance to hardware shop = 5 miles

Time to reach hardware shop = 10 minutes

Time spent at the shop = 30 minutes

Average speed to customer home = 45 mph

Time taken for the travel = 20 minutes

Unknown:

Average speed of the contractor = ?

Solution:

 Average speed is the total distance covered divided by the total time taken.

   Average speed = \frac{total distance}{total time taken}  

     total distance = distance to hardware shop + distance to customer's home

We do not know the distance to customer's home but we have been given the speed and time, so we can find the distance.

  Distance  = speed x time

 Convert the time to hrs and solve;  

                       60 minutes  = 1 hr

                       20 minutes  = \frac{20}{60} hr  = \frac{1}{3} hr

So, Distance  = 45mph x \frac{1}{3} hr   = 15miles

Now;

   Total distance  = 5 + 15 = 20miles

Total time = time to reach hardware shop + time to reach customer's house

                  = 10 + 20

                  = 30 minutes

Convert the time from minutes to hrs;

                 60 minutes  = 1hr

                 30 minutes  = \frac{30}{60}   = 0.5hr

So;

    Average speed  = \frac{20}{0.5} = 40mph

The average speed is 40mph

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Explanation:

Given:

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a.

Time taken by the car to stop:

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b.

using the eq. of motion for the given condition:

v_2^2=u_2^2+2.a_2.\Delta s

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a_x = \frac{d^2}{dt^2}(-19 + t - 3t^3)

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