Two runners are in a rare. Both runners start at the same time. The first runner
ID: 1593111 • Letter: T
Question
Two runners are in a rare. Both runners start at the same time. The first runner travels at a constant velocity of 8 m/s, while the second accelerates from rest at 5 m/s^2 to reach a maximum speed 10 m/s before continuing on at 10 m/s. How long does it take the second runner to catch up to the first runner? How far have t hey each travelled? Joseph can throw a ball vertically with a speed of 15 m/s. How high will it reach? Steph can throw a ball vertically with an initial velocity that is 1.4 times faster than Joseph. How much higher will Steph throw the ball?Explanation / Answer
1) Here total time = t
=> 8t = 10 + 10(t -2)
=> 8t = 10 + 10t - 20
=> 2t = 10
=> t = 5 sec -----------------> time taken by second runner to catch first .
Total distance travelled by each = 8 *5
= 40 m
2) a) height reaches by ball = (15 *15)/(2*9.8)
= 11.48 m
b) height reached by ball for steph = (21 *21)/(2*9.8)
= 22.5 m
=> steph ball reach higher than Joseph by = 22.5 - 11.48
= 11.02 m
3) magnitude of v = sqrt(3.742 + 6.282)
= 7.31
Direction = 180 - tan-1(6.28/3.74)
= 120.77 degrees anticlockwise from + x axis .
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