Asked by anonymous
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
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
Answers
Answered by
oobleck
Recall that starting from rest, s = 1/2 at^2
so,
a = 2s/t^2
v = at = 2s/t
so,
a = 2s/t^2
v = at = 2s/t
Answered by
4TheLoveOfPhysics
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]
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]
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