A car traveling at a speed of vo = 50 m/s stops smoothly (that is, its deceleration is constant) over a distan

Question

a) What is its acceleration during the time it is stopping? (Be careful about the sign!)

Can anyone help me with this

Sorry I wrote the question wrong, the real question is: A car traveling at a speed of vo = 50 m/s stops smoothly (that is, its deceleration is constant) over a distance of d = 157 m.

What is its acceleration during the time it is stopping? (Be careful about the sign!)

Answer 1

[v(f) - v(0) ] / t = a

a = [ 0 - 50 ] / t

a = -50/t , decceleration, since t > 0

Answer 2

Can't be done with what you've given.

v = u + at or v^2 = u^2 + 2as are two equations that might be helpful...if you had the numbers to plug in. But the only thing you gave was u = vo = 50 m/sec. We need either the time (t) the deceleration was smoothly applies to reach final velocity v = 0 or the distance (s) the car stopped in while decelerating.

Reread the problem and look for t or s; or some way to derive one of these.

OK, then, d = s = 157 m. Plug all that into v^2 = u^2 + 2as and solve for a. v = 0, the final velocity, u = 50 m/sec the initial vel., and s = 157 m, the stopping distance. You can do the math.

Lesson Learned:

You need to learn the SUVAT equations that relate velocity, distance, acceleration, and time. Look up SUVAT on the web, you'll find a lot of site explaining what the equations are and how they should be used.

Answer 3

Well you do not know time, and you do not know distance. So you cannot really solve this.

You do know that:
Vf = 0.0m/s
Vo = 50 m/s

You could use this equation:
Vf² = Vo² + 2ax
0 = Vo² + 2ax
-Vo²/2x = a
-2500/2x = a

The accleration will be negative because it is slowing down. The units will be m/s². I hope this helps. You really cannot solve unless you knew time or displacement.