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3 years ago

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A bicyclist makes a trip that consists of three parts, each in the same direction (due north) along a straight road. During the

first part, she rides for 22 minutes at an average speed of 7.2 m/s. During the second part, she rides for 36 minutes at an average speed of 5.1 m/s. Finally, during the third part, she rides for 8.0 minutes at an average speed of 13 m/s. (a) How far has the bicyclist traveled during the entire trip? (b) What is her average velocity for the trip? (Ans: a)26.7km, b) 6.74m/s)

1 answer:

Advocard [28]3 years ago

8 0

Answer:

a

The total distance traveled is $D = 26760 \ m$

b

The average velocity is $v_{avg} = 6.8 \ m/s$

Explanation:

From the question we are told that

The time taken for first part $t_1 = 22 \ minutes = 22*60 = 1320 \ s$

The speed for the first part is $v_1 = 7.2 \ m/s$

The time taken for second part is $t_2 = 36 \ minutes = 2160 \ s$

The speed for the second part is $v_2 = 5.1 \ m/s$

The time taken for third part is $t_3 = 8 \ minutes = 480 \ s$

The speed for the third part is $v_3 = 13 m/s$

Generally

$distance(D) = velocity * time$

Therefore the total distance traveled is

$D = (v_1 * t_1) + (v_2 * t_2 ) + (v_3 * t_3)$

substituting values

$D = (7.2 * 1320) + (5.1 * 2160) + (13 * 480)$

$D = 26760 \ m$

Generally the average velocity is mathematically represented as

$v_{avg} = \frac{D}{t_{total}}$

Where $t_{total}$ is the total time taken which is mathematically represented as

$t_{total} = 1320 + 2160 + 480$

$t_{total} =3960\ s$

The average velocity is

$v_{avg} = \frac{26760}{3960}$

$v_{avg} = 6.8 \ m/s$

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