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A student is running at her top speed of 5.5 m/s to catch a bus, which is stopped at the bus stop. When the student is still a distance 40.4 m from the bus, it starts to pull away, moving with a constant acceleration of 0.173 m/s^2.
(a) For how much time does the student have to run at 5.5 m/s before she overtakes the bus?
(b) For what distance does the student have to run at 5.5 m/s before she overtakes the bus?
(c) When she reaches the bus, how fast is the bus traveling?
(d) If the student's top speed is 2.70 m/s , will she catch the bus?

Respuesta :

onyebuchinnaji

(a) The time taken before the student catches the bus is 11.52 seconds.

(b) The distance the student have to run before she overtakes the bus is 63.36 m.

(c) The speed of the bus is 2.0 m/s.

(d) If the student's top speed is 2.7 m/s he will still catch the bus, since the bus's speed is 2.0 m/s.

Time elapsed before the student over takes the bus

Apply the principle of relative velocity as follows;

(Vs - Vb)t = D

where;

  • Vs is the velocity of the student
  • Vb is the velocity of the bus
  • t is the time in which the student catches the bus
  • D is the distance between them

(Vs - at)t = D

where;

  • a is the acceleration of the bus

(5.5 - 0.173t)t = 40.4

5.5t - 0.173t² = 40.4

0.173t² - 5.5t + 40.4 = 0

solve the quadratic equation using formula method;

a = 0.173, b = -5.5, c = 40.4

t = 11.52 seconds or 20.27 seconds

so the minimum time taken by the student to over take the bus is 11.52 seconds

Distance the student have to run to overtake the bus

d = vt

d = 5.5 m/s x 11.52 s = 63.36 m

Speed of the bus

v = at²

v = 0.173 x (11.52)

v = 2.0 m/s

If the student's top speed is 2.7 m/s he will still catch the bus, since the bus's speed is 2.0 m/s.

Learn more about speed of bus here: https://brainly.com/question/647176

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