Difficult Physics Question to solve!!! Please Help!!?

Question

Please help me with the following question... It is very urgent.. Thank you and please xplain step by step for the best answer.....

A car traveled 50 km in the North direction from Town X towards Town Y. The car left Town X at 10.15 am, stopped for a 10 minute rest along the way and reached town Y at 11.25 am. The car then left for Town Z immediately in the West direction, traveling at 60km/h for 2.0 hours.

Determine the:

(a) straight line distance of the car away from the starting point (also known as displacement),

(b) the mean straight line distance traveled per unit time of the car.


Please help me with this question!!! It's very important! Please...... Thanks a lot :)

? · Accepted Answer

time the car traveled from town X to Y= 11.25 - 10.15= 1 hr 10 min - 10 min(as it stopped) = 1 hr
distance from town X to Y = 50 km

time traveled from town Y to Z= 2 hr
distance from town Y to Z = speed * time = 60 * 2 = 120 km

a) straight line distance of the car away from the starting point using pythagorus theorem
x^2 = 50^2 + 120^2
x^2 = 2500 + 14400
x^2 = 16900
x = sqrt 16900
x= 130 km

b) the mean straight line distance traveled per unit time of the car means the mean speed
speed = displacement/time
speed =  130 km/ 3.166 hr (3 hr 10 min)
speed = 41.06 km/hr

Knowall · Answer

You did not need a help if you had read it carefully.
 deducting the 10 miute break, the car traveled for one hour.
 so it moved 50Km North and reached Y. Then it moved 2 x60= 120 Km West to reach Z.

 a) The straight line distance using pythagorus will be Sqrt((50)^2+(120)^2)= Sqrt (2500+14400) = 130 Km.
 b) The mean str. line speed will be 130Km/(70+120)min. or 130/190= 684 m/minute= 41Km/Hr.

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Difficult Physics Question to solve!!! Please Help!!?

2 Answers