Question

Question 17

Answer 1

So, the acceleration of the car is -1.25 m/s². In other word, the car is also decelerating by 1.25 m/s².

Introduction

Hi ! I will help you to discuss about "deceleration in a straight line movement". Please note in advance that deceleration is acceleration which has a negative value. When an object decelerates, the object will continue to move until it reaches a certain speed (which is less than before) or until it stops. The higher the deceleration value, an object that is moving will stop faster and cover a shorter distance.

Formula Used

In this opportunity, I will give you the following equation to express the relationship between final velocity and initial velocity, acceleration, and distance.

$\boxed{\sf{\bold{(v_t)^2= (v_0)^2 + 2 \times a \times s}}}$

With the following condition:

• $\sf{v_t}$ = final velocity of an object (m/s)
• $\sf{v_0}$ = initial velocity of an object (m/s)
• a = acceleration that happen (m/s²)
• s = the shift or distance of the object (m)

Problem Solving

We know that:

• $\sf{v_0}$ = initial velocity of an object = 72 km/h = 20 m/s
• $\sf{v_t}$ = final velocity of an object = 54 km/h = 15 m/s
• s = the shift or distance of the object = 70 m

Note :

• 1 m/s = 3.6 km/h. So 10 m/s = 36 km/h

What was asked ?

• a = acceleration that happen = ... m/s²

Step by step :

$\sf{(v_t)^2 = (v_0)^2 + 2 \times a \times s}$

$\sf{15^2 = 20^2 + 2 \times a \times 70}$

$\sf{225 = 400 + 140 \times a}$

$\sf{140 a = -175}$

$\sf{a = \frac{-175}{140}}$

$\boxed{\sf{\bold{a = -1.25 \: m/s^2}}}$

Conclusion

Here, we see that the acceleration is -1.25 m/s². In other words, the car is also decelerating by 1.25 m/s².