A car enters the freeway with a speed of 6.4 m/s and accelerates uniformly for 3.2 km in 3.5 min. How fast is the car moving after this time.
Thanks!
Thanks!
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t₀ = 0, x₀ = 0, v₀ = 6.4 m/s
t₁ = 3.5 min = 210 s, x₁ = 3.2 km = 3200 m, v₁ = ?
a = constant = ?
Two unknowns; we will need two equations.
When acceleration is constant:
(I) ∆v = a∆t; v₁ - v₀ = a(t₁ - t₀) = at₁
(II) v₁² - v₀² = 2a∆x = 2ax₁
Divide (II) by (I):
(III) v₁ + v₀ = 2x₁/t₁
v₁ = 2x₁/t₁ - v₀ = (6400 m / 210 s) - 6.4 m/s
= 24.1 m/s
[Note that (III) can be obtained more immediately by observing that, when acceleration is constant, the average speed is just the total distance traveled, divided by the total time taken.
Or you can get (II) by using the fact that:
change in kinetic energy = work done = force•distance, then divide by the mass:
∆(½mv²) = F∆x = ma∆x
½(v₁² - v₀²) = a∆x
]
t₁ = 3.5 min = 210 s, x₁ = 3.2 km = 3200 m, v₁ = ?
a = constant = ?
Two unknowns; we will need two equations.
When acceleration is constant:
(I) ∆v = a∆t; v₁ - v₀ = a(t₁ - t₀) = at₁
(II) v₁² - v₀² = 2a∆x = 2ax₁
Divide (II) by (I):
(III) v₁ + v₀ = 2x₁/t₁
v₁ = 2x₁/t₁ - v₀ = (6400 m / 210 s) - 6.4 m/s
= 24.1 m/s
[Note that (III) can be obtained more immediately by observing that, when acceleration is constant, the average speed is just the total distance traveled, divided by the total time taken.
Or you can get (II) by using the fact that:
change in kinetic energy = work done = force•distance, then divide by the mass:
∆(½mv²) = F∆x = ma∆x
½(v₁² - v₀²) = a∆x
]
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Thanks for the BA!
Check out akm69's answer. That's yet another good way to do it.
He solves the distance equation for the acceleration, then plugs that back into the velocity equation to get the final velocity.
[TU!]
Check out akm69's answer. That's yet another good way to do it.
He solves the distance equation for the acceleration, then plugs that back into the velocity equation to get the final velocity.
[TU!]
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A small point, perhaps, but worth noting --
It is often advantageous to work everything through symbolically, not plugging any numerical values in until the end.
It is often advantageous to work everything through symbolically, not plugging any numerical values in until the end.
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By s = ut + 1/2at^2
=>3200 = 6.4 x (3.5 x 60) + 1/2 x a x (3.5 x 60)^2
=>a = 8.42 x 10^-2 m/s^2
BY v = u + at
=>v = 6.4 + 8.42 x 10^-2 x (3.5 x 60)
=>v = 24.08 m/s
=>3200 = 6.4 x (3.5 x 60) + 1/2 x a x (3.5 x 60)^2
=>a = 8.42 x 10^-2 m/s^2
BY v = u + at
=>v = 6.4 + 8.42 x 10^-2 x (3.5 x 60)
=>v = 24.08 m/s