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myrzilka [38]
3 years ago
9

Issac and Blaise decide to race. They both start at the same position at the same time. Issac runs at 2m/s but decides to take a

2 minute rest stop along the way. Blaise runs at half the speed and still wins by 10 m. How far did Blaise run? Please show your work, it’ll greatly help with my understanding. I know I have to do some substitution with equations, but I don’t have a final length of the race, or final time, so I have no idea where to start.
Physics
1 answer:
FromTheMoon [43]3 years ago
6 0

Let the Blaise runs for time "t" to complete the race

so the total distance he moved is given by

d_1 = 1* t

Now Issac runs for time t = "t - 2*60"

because it took rest for 2 minutes

d_2 = 2*(t - 120)

now it is given that Blaise wins by 10 m distance

d_1 - d_2 = 10

1* t - 2*(t - 120) = 10

t - 2t + 240 = 10

t = 230 s

now the distance moved by Blaise is given by

d_1 = 1*230 = 230 m

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1. This question asks about velocity, so A and B are not correct. The car's velocity after 15 s with acceleration 2.00 m/s² would be

(2.00 m/s²) (15 s) = 30 m/s

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2. Because Ima is slowing down to a stop, the acceleration is negative. Let <em>x</em> be the displacement of her vehicle during this motion. Then

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If <em>x</em> is the distance traveled as the car comes to a stop, then

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4.a. Assuming the sprinter's acceleration is constant, the average acceleration would be <em>a</em> such that

(11.5 m/s)² - 0² = 2 <em>a</em> (15.0 m)

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4.b. By definition of average acceleration,

4.41 m/s² = (11.5 m/s - 0 m/s)/<em>t</em>

==>   <em>t</em> = (11.5 m/s)/(4.41 m/s²) ≈ 2.61 s

5. At maximum height, any thrown object has zero velocity, so if it was thrown with an initial speed <em>v</em>, at its highest point we have

0² - <em>v</em> ² = 2 (-<em>g</em>) (91.5 m)

==>   <em>v</em> = √(2<em>g</em> (91.5 m)) ≈ 42.3 m/s

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6.b. The brick is presumably dropped from rest, so it is displaced by <em>x</em> such that

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