1. A football player initially at rest accelerates uniformly as she runs down the field, travelling 17 m [E] in 3.8 s. What is her final velocity?
2. A child on a toboggan sits at rest on the top of a tobogganing hill. If the child travels 70.0 m [downhill] in 5.3 s while accelerating uniformly, what acceleration does the child experience
Question 1)
vi = 0 m/s,
Δd = 17 m [E],
Δt = 3.8 s,
a = ?
Δd = vi(Δt) + 1/2(a)(Δt²)
Δd = 0(Δt) + 1/2(a)(Δt²)
Δd = 1/2(a)(Δt²)
a = 2Δd / Δt²
a = 2(17) / (3.8)²
∴a = 2.4 m/s² [E]
Question 2)
vi = 0 m/s,
Δd = 70.0 m [downhill],
Δt = 5.3 s,
a = ?
Δd = vi(Δt) + 1/2(a)(Δt²)
Δd = 0(Δt) + 1/2(a)(Δt²)
Δd = 1/2(a)(Δt²)
a = 2Δd / Δt²
a = 2(70.0) / (5.3)²
∴a = 5.0 m/s² [downhill]
Recall that starting from rest, s = 1/2 at^2
so,
a = 2s/t^2
v = at = 2s/t
1. Well, if she's running down the field, she must really be in a hurry to get to the end zone! To find her final velocity, we can use the formula v = u + at, where v is the final velocity, u is the initial velocity (which is 0 since she starts from rest), a is the acceleration, and t is the time. Plugging in the values, we have v = 0 + (17 / 3.8) = 4.47 m/s [E]. Looks like she's running pretty fast!
2. Ah, the thrill of tobogganing down a hill! The child must be having a slippery and slidey adventure. Now, to find the acceleration experienced by the child, we can use the formula s = ut + (1/2)at^2, where s is the distance traveled, u is the initial velocity (which is 0 since the child starts at rest), a is the acceleration, and t is the time. Plugging in the values, we have 70 = 0 + (1/2)a(5.3^2). Solving this equation, we find that the child experienced an acceleration of approximately 3.35 m/s^2 [downhill]. Wheee!
To answer the first question:
1. We can use the formula for uniform acceleration: vf = vi + at, where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time.
2. In this case, the football player starts at rest, so the initial velocity (vi) is 0 m/s.
3. The distance traveled (d) is given as 17 m, and the time (t) is given as 3.8 s.
4. We need to rearrange the formula to solve for vf: vf = (d - vi) / t = d / t.
5. Plugging in the values, vf = 17 m / 3.8 s ≈ 4.47 m/s.
Therefore, the football player's final velocity is approximately 4.47 m/s [E].
To answer the second question:
1. We can again use the formula vf = vi + at, where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time.
2. In this case, the child starts at rest, so the initial velocity (vi) is 0 m/s.
3. The distance traveled (d) is given as 70.0 m, and the time (t) is given as 5.3 s.
4. Rearranging the formula to solve for a, we get a = (vf - vi) / t = vf / t (since vi = 0).
5. Plugging in the values, a = 70.0 m / 5.3 s ≈ 13.21 m/s².
Therefore, the child experiences an acceleration of approximately 13.21 m/s² downhill.