So, the acceleration of the car is -1.25 m/s². In other word, the car is also decelerating by 1.25 m/s².
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.
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:
We know that:
Note :
What was asked ?
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]
Here, we see that the acceleration is -1.25 m/s². In other words, the car is also decelerating by 1.25 m/s².
