n a 100 m race, a runner Nathan accelerating uniformly takes 2.0 s and another runner Sammy 2.5 s to reach their maximum speeds which they each maintain for the rest of the race. They cross the finish line simultaneously, both setting a time of 12.0 s. (a) What is the acceleration of each runner? (b) What are the respective maximum speeds of Nathan and Sammy? (c) Which runner is ahead at the 6.0 s, and how much? (d) Sketch the speed – time graphs for the runners in same axis.
Answers
Given : 100 m race
a runner Nathan accelerating uniformly takes 2.0 s and another runner Sammy 2.5 s to reach their maximum speeds which they each maintain for the rest of the race.
Both finish race in 12 secs
To find :
(a) What is the acceleration of each runner?
(b) What are the respective maximum speeds of Nathan and Sammy?
(c) Which runner is ahead at the 6.0 s, and how much?
(d) Sketch the speed – time graphs for the runners in same axis.
Solution:
Let say acceleration of Nathan = a₁ m/s²
v = u + at
Max speed = 0 + a₁(2) = 2a₁ m/s
Distance covered in 1st 2sec = (1/2)(0 + 2a₁ ) 2 = 2a₁ m
Distance covered in remaining 10 secs = 2a₁ * 10 = 20a₁ m
2a₁ + 20a₁ = 100
=> a₁ = 100/22 = 50/11 = 4.55 m/s²
acceleration of Nathan = 4.55 m/s²
max speed of Nathan = 2 * 4.55 = 9.1 m/s
Nathan Distance in 6 secs = 2a₁ + 4(2a₁) = 10a₁ = 45.5 m
Let say acceleration of Sammy = a₂ m/s²
v = u + at
Max speed = 0 + a₂(2.5) = 2.5a₂ m/s
Distance covered in 1st 2.5sec = (1/2)(0 + 2.5a₂ ) 2.5 = 3.125a₂ m
Distance covered in remaining 9.5 secs = 2.5a₂ * 9.5 = 23.75a₂ m
3.125a₂ + 23.75a₂ = 100
=> a₂ = 100/26.875 = 3.72 m/s²
acceleration of Sammy = 3.72 m/s²
max speed of Sammy = 2.5 * 3.72 = 9.3 m/s
Sammy Distance in 6 secs = 3.125a₂+ 3.5(2.5a₂) = 11.875a₂ = 44.2 m
Hence Nathan is ahead at 6 secs by 45.5 - 44.2= 1.3 m
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