The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with an acceleration of -5.6m/s for 4.20s marking straight skid marks 62.4m long ending at the tree with what speed does the car strike the tree?
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To find the speed at which the car strikes the tree, we can use the equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity (unknown)
u = initial velocity (assumed to be 0 since the car initially comes to a stop)
a = acceleration (-5.6 m/s^2, negative because it's deceleration)
s = distance (62.4 m)
Substituting the given values into the equation and solving for v:
v^2 = 0 + 2(-5.6)(62.4)
v^2 = -698.88
v = √(-698.88) (taking the square root of both sides of the equation)
Since the square root of a negative number is undefined in real numbers, it means that the car does not strike the tree with a positive velocity. Instead, it comes to a stop just before reaching the tree.
To find the speed at which the car strikes the tree, we can use the equation of motion:
v^2 = u^2 + 2as
where:
v = final velocity (which we want to find)
u = initial velocity (assumed to be zero as the car started from rest)
a = acceleration (-5.6 m/s^2)
s = distance (62.4 m)
First, let's rearrange the equation to solve for v:
v^2 = 2as - u^2
v^2 = 2(-5.6)(62.4) - 0
v^2 = -2(-5.6)(62.4)
v^2 = 699.84
v = √699.84
v ≈ 26.46 m/s
So, the speed at which the car strikes the tree is approximately 26.46 m/s.