Help with acceleration and velocity?

Question

I really do not get this problem. It's my first week in physics and I am dying already.

A car moving at constant speed travels 110 meters in 5.0 seconds. If it then brakes and comes to a stop in 4.0 seconds, what is its acceleration in m/s^2?

Answer 1

There are four basic equations of motion for an object moving with constant acceleration:

d = x + ut + ½at²

d = ½(u + v)t

v = u + at

v² = u² + 2ad

where

...a = acceleration

...d = distance traveled during time t

...u = initial velocity at time, t = 0

...v = velocity after time t

...x = initial position at time, t = 0

You can derive most other motion equations from these basic four. Solving motion problem will become easier once you get used to them.

As the 1st answer gave, you need to:

1. Compute the initial speed:

... u = d/t = 110 m/5.0 s

2. Compute the acceleration:

... v = u + at

... 0 = u + at

... a = -u/t = -(110 m/5.0 s)/4.0 s

When given 2 or more times, distances, speed, whatever, be careful which you use where.

Answer 2

first: find the constant speed, 110/5 = 22m/s , that'll also be you initial velocity.

ur final velocity is 0, since the car eventually stopped.

vf = v i + at

0 = 22 + (a) 4

a = - 5.5 m/s^2.

this is a simple problem, try asking ur teacher for help, because the course will get more complicated than this soon.

Answer 3

I think in your text you will find a formulae something like A = (u-v) / t , where

A = acceleration

u = final speed

v = initial speed, and

t = time



This means Acceleration = the change in speed divided by the time.

The initial velocity = meters / second = 110/5 =  22

The final velocity = 0

So the change in velocity =   -22 m/s (negative, since it was slowing down).

The time the speed change took was 4 seconds

So the acceleration = change in velocity / time = -22 / 4 = 5.5 m/s^2

Note m/s^2 is the same as m/s/s (meters per second per second)

Answer 4

wow, I totally forgot how to do this and I'm taking physics this year again.  ahhh.