Question

Starting from rest a car accelerates at 2.0m/s2 up a hill that is inclined 5.5 degree above the horizontal. How far (a) horizontally and (b)vertically has the car traveled in 12seconds?

Answer 1

Given:

Acceleration = 2.0 m/s²

θ = 5.5 degrees above the horizontal.

Let's solve for the following:

(a) Hwo far horizontally has the car travelled in 12 seconds.

To find the horizontal distance, apply the formula:

$d_x=d\cos \theta$

To solve for d, apply the motion formula:

$d=V_it+\frac{1}{2}at^2$

Where:

Vi is the initial velocity = 0 m/s

t is the time = 12 seconds

a is the acceleration = 2.0 m/s²

Thus, we have:

$\begin{gathered} d=0(12)+\frac{1}{2}\ast2.0\ast12^2 \\ \\ d=0+144 \\ \\ d=144\text{ m} \end{gathered}$

Thus, to find the horizontal distance, we have:

$\begin{gathered} d_x=d\cos \theta \\ \\ d_x=144\cos 5.5 \\ \\ d_x=143.3\text{ m} \end{gathered}$

The car traveled 143.3 meters horizontally in 12 seconds.

• (b) To find the vertical distance in 12 seconds, apply the formula:

$d_y=d\sin \theta$

Thus, we have:

$\begin{gathered} d_y=144\sin 5.5 \\ \\ d_y=13.8\text{ m} \end{gathered}$

The car traveled 13.8 meters vertically in 12 seconds.

ANSWER:

(a) 143.3 m

(b) 13.8 m