Question

URGENT

A student runs at 4.5 m/s [27° S of W] for 3.0 minutes and then he turns and runs at 3.5 m/s [35° S of E] for 4.1 minutes.

a. What was his average speed?

b. What was his displacement?

PLEASE SHOW ALL WORK​

Answer 1

(a) 3.93 m/s

(b) 861.66 m

Step-by-step explanation:

A = 4.5 m/s [27° S of W] for 3.0 minutes

B = 3.5 m/s [35° S of E] for 4.1 minutes

Distance A = 4.5 x 3 x 60 = 810 m

Distance B = 3.5 x 4.1 x 60 = 861 m

(a) The average speed is defined as the ratio of the total distance to the total time.

Total distance, d = 810 + 861 = 1671 m

total time, t = 3 + 4.1 = 7.1 minutes = 7.1 x 60 = 426 seconds

The average speed is

`v=(1671)/(426)=3.93 m/s`

(b)

`\overrightarrow{A} = 810(- cos 27 \widehat{i} - sin 27 \widehat{j})=- 721.7 \widehat{i} - 367.7 \widehat{j}\\\\\overrightarrow{B} = 861( cos 35 \widehat{i} - sin 35 \widehat{j})= 705.3 \widehat{i} - 493.8 \widehat{j}\\\\\overrightarrow{C} = (- 721.7 + 705.3) \widehat{i} - (367.7 + 493.8) \widehat{j} \\\\\overrightarrow{C}= - 16.4 \widehat{i} - 861.5 \widehat{j}`

The magnitude is

`C =√(16.4^2+861.5^2) = 861.66 m`