Answer :
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.
[tex] \boxed{\sf{\bold{(v_t)^2= (v_0)^2 + 2 \times a \times s}}}[/tex]
With the following condition:
- [tex] \sf{v_t} [/tex] = final velocity of an object (m/s)
- [tex] \sf{v_0} [/tex] = 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:
- [tex] \sf{v_0} [/tex] = initial velocity of an object = 72 km/h = 20 m/s
- [tex] \sf{v_t} [/tex] = 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 :
[tex] \sf{(v_t)^2 = (v_0)^2 + 2 \times a \times s} [/tex]
[tex] \sf{15^2 = 20^2 + 2 \times a \times 70} [/tex]
[tex] \sf{225 = 400 + 140 \times a} [/tex]
[tex] \sf{140 a = -175} [/tex]
[tex] \sf{a = \frac{-175}{140}} [/tex]
[tex] \boxed{\sf{\bold{a = -1.25 \: m/s^2}}} [/tex]
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.