A hockey stick is in contact with a 165-g puck for 22.4 ms; during this time, the force on the puck is given approximately by f(
t)=a+bt+ct2, where a =
1 answer:
I've found the complete form of this problem from another website which is shown in the attached picture. The equation is:
F(t) = -25 + (1.23×10⁵)t - (5.58×10⁶)t²
a.) For the first question, let's substitute t = 22.4×10⁻³ seconds to the formula.
F = -25 + (1.23×10⁵)(22.4×10⁻³ s) - (5.58×10⁶)(22.4×10⁻³ s)²
F = -69.62 N
From Newton's second law, F = ma.
-69.62 = (165 g)(1 kg/1000 g)(a)
Solving for a,
a = -421.94 m/s² = Δv/Δt = (0 - v)/(22.4×10⁻³ - 0)
Solving for v,
<em>v = 9.45 m/s</em>
b.) To solve for the distance, the formula is:
d = v₀t + 1/2(a)(t²)
Let's use the absolute value of a because distance is always positive.
d = 0(22.4×10⁻³) + 1/2(421.94)(22.4×10⁻³)²
<em>d = 0.106 m</em>
F(t) = -25 + (1.23×10⁵)t - (5.58×10⁶)t²
a.) For the first question, let's substitute t = 22.4×10⁻³ seconds to the formula.
F = -25 + (1.23×10⁵)(22.4×10⁻³ s) - (5.58×10⁶)(22.4×10⁻³ s)²
F = -69.62 N
From Newton's second law, F = ma.
-69.62 = (165 g)(1 kg/1000 g)(a)
Solving for a,
a = -421.94 m/s² = Δv/Δt = (0 - v)/(22.4×10⁻³ - 0)
Solving for v,
<em>v = 9.45 m/s</em>
b.) To solve for the distance, the formula is:
d = v₀t + 1/2(a)(t²)
Let's use the absolute value of a because distance is always positive.
d = 0(22.4×10⁻³) + 1/2(421.94)(22.4×10⁻³)²
<em>d = 0.106 m</em>
You might be interested in