Answer:
Explanation:
1) The number of times, the car with the speed of 195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.
0 .5*195/2.5 = 39
Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times
and car with the speed of 180 mph crosses it 36 times
here, the time-mean speed, vt is given below,
vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)
= 186.433 mph
and space mean speed is given by,
= (39+38+37+36)/(39/195+38/190+37/1850+36/180)
1) The number of times, the car with the speed of 195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.
0 .5*195/2.5 = 39
Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times
and car with the speed of 180 mph crosses it 36 times
here, the time-mean speed, vt is given below,
vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)
= 186.433 mph
and space mean speed is given by,
= (39+38+37+36)/(39/195+38/190+37/1850+36/180)
=187.5 mph
2) There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt in that condition will be calculated as
Vt = 195+190+185+180/4
= 187.5
Vs= 4/(1/195+1/190+1/185+1/180)
= 188.36 mph
2) There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt, in that condition will be calculated as
Vt = 195+190+185+180/4
= 187.5
Vs= 4/(1/195+1/190+1/185+1/180)
= 188.36 mph
