Question

A truck initially at rest moves down a steep slope in a straight line at a rate of 5 ms-2 for 4 s. How far does the truck during this time?

Answer 1

Answer:

40 m

Explanation:

As per the provided information in the given question, we have :

• Initial velocity (u) = 0 m/s (As the truck was at rest)
• Acceleration (a) = 5 m/s²
• Time taken (t) = 4 seconds

We are asked to calculate the distance travelled or how far does the truck travel during this time.

Here, we'll be using a equation of motion from the 3 equations of motion. Take a look at the formulae given below :

$\boxed{ \begin{array}{cc} \pmb{\sf{ \quad \: v = u + at \quad}} \\ \\ \pmb{\sf{ \quad \: s= ut + \cfrac{1}{2}a{t}^{2} \quad \: } } \\ \\ \pmb{\sf{ \quad \: {v}^{2} - {u}^{2} = 2as \quad \:}}\end{array}}$

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• t denotes time
• s denotes distance

Only from the 2nd and 3rd equations we can find the distance travelled. Here we've u,a,t , so by using the second equation of motion, we san easily find the value of distance.

By using the second equation of motion :

⇒ s = ut + ½at²

⇒ s = 0(4) + ½ × 5 × (4)²

⇒ s = ½ × 5 × 16

⇒ s = 5 × 8

⇒ s = 40 m

∴ The truck travels 40 m during this time.

$\rule{200}2$