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?
Vo*t + 0.5a*t^2 = 62.4 m.
Vo*4.2 - 0.5*5.6*4.2^2 = 62.4,
4.2Vo - 49.39 = 62.4,
4.2Vo = 111.79, Vo = 26.6 m/s.
V^2 = Vo^2 + 2a*d. Vo = 26.6 m/s, a = -5.6 m/s^2, d = 62.4 m,
V = ?.
^^ he's right thx bud
To find the speed at which the car strikes the tree, we can use the kinematic equation:
v^2 = u^2 + 2as
where:
v = final velocity (unknown)
u = initial velocity (0 m/s since the car starts from rest)
a = acceleration (-5.6 m/s^2)
s = distance (62.4 m)
Plugging in the values into the equation, we can solve for v:
v^2 = (0 m/s)^2 + 2(-5.6 m/s^2)(62.4 m)
v^2 = 0 + (-11.2 m/s^2)(62.4 m)
v^2 = -699.84 m^2/s^2
Since velocity cannot be negative in this context, we take the positive square root:
v = √(-699.84 m^2/s^2)
v ≈ 26.45 m/s
Therefore, the car strikes the tree with a speed of approximately 26.45 m/s.
To find the speed at which the car strikes the tree, we need to use the kinematic equation that relates displacement, initial velocity, acceleration, and time. The equation is:
\[ v^2 = u^2 + 2as \]
Where:
v = final velocity (which is the speed at which the car strikes the tree)
u = initial velocity (which is 0 because the car is coming to a stop)
a = acceleration (-5.6m/s²)
s = displacement (62.4m)
Let's solve the equation step by step:
1. Substitute the known values into the equation:
\[ v^2 = 0^2 + 2(-5.6)(62.4) \]
2. Simplify the equation:
\[ v^2 = -2(5.6)(62.4) \]
3. Calculate the right side of the equation:
\[ v^2 = -698.88 \]
4. Take the square root of both sides to find v:
\[ v = \sqrt{-698.88} \]
Note: The answer will have both positive and negative values. However, since speed cannot be negative, we will ignore the negative sign.
5. Calculate the square root:
\[ v = 26.43 \, \text{m/s} \]
Therefore, the car strikes the tree with a speed of approximately 26.43 m/s.