# Homework Help: Speed, Acceleration and Time - SUVAT Equations and Speed Time Graph

1. Jun 11, 2012

### mm391

1. The problem statement, all variables and given/known data

In 4 seconds a car travels with constant velocity of 50m/s. After 4 seconds the car decelerates at 10m/s^2 until velocity = 0. What is the velocity at 8 and 12 seconds and what is the distance traveled after 12 seconds?

Also how would I draw the acceleration time graph and relate that to a speed time graph?

2. Relevant equations

f(t)=U(t)+a(t^2/2)=s

3. The attempt at a solution

But this does not seem to give me the correct answer. I was told the total distance is 370 meters

2. Jun 11, 2012

### zaal

8 seconds total is 4 seconds after the deceleration starts. So the velocity at t=8s is equal to 50 m/s - 4 s ( 10 m/s2 ) = 10 m/s

12-4=8 so 50 m/s - 8 s ( 10 m/s2 ) = -30 m/s, however the deceleration end at v= 0 m/s. So the speed actually is 0 m/s.

To find the total distance traveled one has to find to the moment the car first has speed v=0. To do that solve the equation: 50 m/s - tstop ( 10 m/s2 ) = 0
Solving gives tstop = 5 s (actually 5 s for the moment deceleration so 9 seconds total)
The average speed traveled is (50 m/s + 0 m/s)/2 = 25 m/s (this work only if the deceleration or accelaration is constant)
The distance traveled during deceleration is 25 m/s * 5 s = 125 meters
The distance traveled before deceleration is 50 m/s * 4 s = 200 meters
The total distance traveled is 325 meters.

An acceleration vs time chart would look like this (x-axis should be t-axis):
http://www4c.wolframalpha.com/Calculate/MSP/MSP20271a1ifb1cc14f2e4i000043e243867d21fe1a?MSPStoreType=image/gif&s=41&w=290&h=123&cdf=RangeControl [Broken]

A speed vs time chart would look like this (x-axis should be t-axis):
http://www4c.wolframalpha.com/Calculate/MSP/MSP731a1ifd56g88acbe20000156h79ga9hi13fed?MSPStoreType=image/gif&s=37&w=290&h=136&cdf=RangeControl [Broken]

A location vs time chart would look like this:
http://www4c.wolframalpha.com/Calculate/MSP/MSP6251a1ifd95dbc17f8b00006842i3gc0ig97h33?MSPStoreType=image/gif&s=26&w=287&h=127&cdf=RangeControl [Broken]

Last edited by a moderator: May 6, 2017
3. Jun 11, 2012

### azizlwl

Zaal post is not correct.
The question is the distance travelled not displacement.
Try to draw the diagram.
At t=4secs, draw line with gradient -10 till t=12sec.
Add the areas(both sides taken as positive, since it is distance covered) and you get the answer.

4. Jun 11, 2012

### zaal

Sorry I made one mistake I corrected it (I forgot to add the distance traveled before deceleration).
PS Don't forget v=0 for t>9

5. Jun 11, 2012

### azizlwl

Your graph should be extended till t=12sec
The distance =4(50)+(50)(50)/20 +1/2(10)(3)2 =200+125+45=370m

6. Jun 11, 2012

### ehild

You ignore the following sentence of the problem:

The wording of the problem is confusing.

ehild