Basic Kinematics: Solving for an Object's Motion with Constant Acceleration

[themilkman]

Homework Statement

An object is out in free space, in motion along a straight line, with a constant acceleration of 15 m/s2.
At time t = 0, the object is at x = 0 meters, with a velocity of -80 m/s. Where is the object at t = 10s ?

a. - 50 m

b. 20 m

c. - 4 m

d. 75 m

Homework Equations

The Attempt at a Solution

I'm not even sure how to start this problem. This class for some reason does not use calculus, and the teacher is terrible so I have no idea what to do. My best guess would be that 70m and the way I tried this was assuming since V= -80 m/s and constant accel - 15m/s2 that t1= -65m and then to add 15 m every second until t10 = 70. Only problem is this is not an answer.

What should I do?

 

[lukas86]

At the Introductory Physics page where you posted this, there is a sticky at the top with different formulas in there. The link is <https://www.physicsforums.com/showthread.php?t=110015> and the formula you need is in there. It's in the second post that Doc Al posted under "General".

You have all the information you need. Starting position, starting velocity, acceleration, and time.

 

[baba89]

I would approach this problem by first understanding the basic principles of kinematics. In this case, we are dealing with an object in motion along a straight line with constant acceleration. This means that the object's velocity is changing at a constant rate of 15 m/s^2, in this case, becoming more negative as time goes on.

To solve for the object's position at t = 10s, we can use the kinematic equation x = x0 + v0t + 1/2at^2, where x0 is the initial position, v0 is the initial velocity, a is the acceleration, and t is the time.

In this problem, we are given x0 = 0 m, v0 = -80 m/s, a = 15 m/s^2, and t = 10 s. Plugging these values into the equation, we get x = (0 m) + (-80 m/s)(10 s) + 1/2(15 m/s^2)(10 s)^2 = -400 m + 750 m = 350 m.

Therefore, at t = 10 s, the object is at a position of 350 m, which is option d.