# A truck starts from rest at the top a slope which is 1 m high and 49 m long. What is the acceleration and it's speed at the bottom of the slope assuming friction is negligible?

Jun 18, 2018

$a = 0.2 \frac{m}{s} ^ 2$
$v = 4.4 \frac{m}{s}$

#### Explanation:

Acceleration:
The truck's weight is m*g. Weight, w, is a force. If the angle between the slope's surface and horizontal is $\theta$, then the component of w that points down the slope is

${w}_{\text{ds}} = m \cdot g \cdot \sin \theta$

It might help to draw a side view of the situation. Since sine is opposite/adjacent,

${w}_{\text{ds}} = m \cdot g \cdot \frac{1 m}{49 m} = \frac{m \cdot g}{49}$

From Newton's 2nd Law

$a = {w}_{\text{ds}} / m = \frac{\frac{\cancel{m} \cdot g}{49}}{\cancel{m}} = \frac{g}{49} = \frac{9.8 \frac{m}{s} ^ 2}{49} = 0.2 \frac{m}{s} ^ 2$

Speed:
Use the formula

${v}^{2} = {u}^{2} + 2 \cdot a \cdot d = 0 + 2 \cdot 0.2 \frac{m}{s} ^ 2 \cdot 49 m = 19.6 {m}^{2} / {s}^{2}$

$v = \sqrt{19.6 {m}^{2} / {s}^{2}} = 4.4 \frac{m}{s}$

I hope this helps,
Steve