Starting from rest, a car accelerates at 2.1 m/s2 up a hill that is inclined 5.1 ∘ above the horizontal. How far horizontally has the car traveled in 14 s ? How far vertically has the car traveled in 14 s ?
Express your answer using two significant figures.
distance up the hill (d) ... 1/2 a t^2 ... 1/2 * 2.1 * 14^2
horizontal ... d * cos(5.1º)
vertical ... d * sin(5.1º)
To find the horizontal distance traveled by the car, we can use the equation:
\[d = v_i t + \frac{1}{2} a t^2\]
Where,
\(d\) is the distance traveled
\(v_i\) is the initial velocity (which is 0 since the car starts from rest)
\(a\) is the acceleration
\(t\) is the time
Plugging in the given values:
\(v_i = 0\)
\(a = 2.1 \, \text{m/s}^2\)
\(t = 14 \, \text{s}\)
We can ignore the first term (0) since the initial velocity is zero, and then substitute in the values for the remaining terms:
\[d = \frac{1}{2} \cdot 2.1 \, \text{m/s}^2 \cdot (14 \, \text{s})^2\]