Tuesday, 16 September 2014

DISPLACEMENT, SPEED , VELOCITY PROBLEMS AND SOLUTIONS


Problem1:
While traveling along a straight interstate highway you notice that the mile marker reads 260.  You travel until you reach the 150-mile marker, and then retrace your path to the 175-mile marker.  What is the magnitude of your resultant displacement from the 260-mile marker?
  • Solution:
    The resultant displacement is the vector d, the sum of two vectors d1 and d2 which point in opposite directions.  The  magnitude of the resultant displacement vector is (260 - 175) miles = 85 miles.

Problem2:

In a time interval of 5 minutes, a runner runs once around a one-mile track.  What is his average velocity?  What is his average speed?
  • Solution:
    After 5 minute the runner returns to his starting position.  The displacement is zero, so his average velocity is zero.
    The average speed is the distance traveled in the time interval Dt.  This distance is one mile.  The average speed therefore is (1 mile)/(5 minutes) = (12 miles)/(60 minutes) = 12 miles/h.
    Note: Speed is a scalar, velocity is a vector.  The average speed is in general not equal to the magnitude of the average velocity.

Problem3:

A motorist drives north for 35 minutes at 85 km/h and then stops for 15 minutes.  He then continues north, traveling 130 km in 2 hours.
(a) What is his total displacement?
(b) What is his average velocity?
  • Solution:
    (a) In the first 35 minutes the motorist travels
    d= v1t= 85 km/h ´ 35 min ´ 1 h/(60 min) = 49.6 km.
    In the next 2 hours he travels 130 km.  The total distance traveled is 179.6 km.  His displacement is 179.6 km (north).
    (b) His average velocity is d/t.  He travels for 170 minutes (including his stop). Therefore his average velocity is  
    = (179.6 km/(170 min)) ´ (60 min/h) (north) = 63.4 km/h (north).