Distance In Softball Ilab Answers

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Faculty of Computer Studies
Physics
TMA
Spring 2014 2015

Name of Student: Hala Saleh Flitti
Signature:
Date:25th of April 2015 Questions of Chapter 3: Question 16: page 111
(Distance in softball game)/(Speed of softball) =(Distance in baseball game)/(Speed of baseball)

(43 ft)/(65 mph) =(60.5 ft)/(Speed of baseball)

Speed of baseball = (60.5 ft)/43ft (65 mph) = 91.5 mph

Question 24: page 112
From 0S to 1S: d1 = 2m

Distance moved in the interval 1S to 3 S: d2= 10 m

Distance moved in the interval 3S to 5S: d3= 15 m

Distance moved in the interval 5S to 6S: d4= 4 m

Distance moved in the interval 6S to 8S: d5= 1m

Therefore the total distance travelled at 8S: D = d1+d2 +d3 + d4 + d5 = 32 m
…show more content…
Distance travelled = x= (24m/s) (9s) + 1/2 (-2m/s2) (9s)2
= 135 m
Question 14: page 148 Time taken by the car to stop can be calculated as follows:
Velocity of car when it stops = speed of car+ (slow down acceleration) time
Thus,
0 = 27 m/s + (-7 m/s2) t
Therefore, t = (27m/s)/(7 m/s^2 ) = 3.85 s The distance covered by the car till it stops (displacement of the car)
= 0+ (27m/s) (3.85s) + 1/2 (-7 m/s2) (3.85s)2 = 103.95m – 51.88m = 52
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