4. A student is running at a constant velocity of 6.00 m/s to catch the campus s

Question

4. A student is running at a constant velocity of 6.00 m/s to catch the campus shuttle bus,
which is stopped at the bus stop. When the student is 80.0 m from the bus, it starts to pull away
with a constant acceleration of 0.200 m/s2.
a) How far does the bus travel in the first 10.0 s?
b) How far has the bus traveled when its speed is 1.75 m/s?
c) From the instant when the bus starts accelerating, how long will the student have to run
before catching the bus?
d) How fast will the bus be traveling when the student catches the bus?
e) The equation you solved in c) has a second solution. What is the significance of that
solution?

Explanation / Answer

a. == 10.0 m b == 7.66m c == 20.0 s or 40.0 s d == 4.00 m/s e == The student first catches up with the bus at t=20.0s. If she keeps running, she then passes the bus. The bus continues to accelerate and passes the student when t = 40.0s f == 5.66 m/s == what i've done. i've used formulas to try to solve these: x = x0 +v0t + 1/2at^2 and ... v = v0 +at and ... v^2 = v0 + 2a*(chng Of X) however, i don't get the right answers, i know to get the right answers i can't mix match the information in the formula. okay for a. and b. i get the correct answers can you please verify my work is correct. and i need help setting up the rest? ... for a, i first found velocity: v = 0 + (.2)(10s) = 2 m/s. then used a different formula to find the distance for questions a. so x = 0 + 0(10) + 1/2(.2)(10^2) = 10 m. ... alright for b. i just used the third formula: (1.75)^2 = 0 + 2(.2)(chgofx) === 7.656

Related Questions

Navigate

• Browse (All)
• Browse 4
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.