Kinematics and dynamics

  • Thread starter Hoejer
  • Start date
  • #1
6
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Homework Statement


I want to calculate the travelled length of a car.

A travelling car with a acceleration from 0 m/s to 19.4 m/s in 10 sec. From there it has a speed of 19.4 m/s for 11 sec. After the 21 sec. the car hit the brakes, and deaccelerates to 0m/s in 3 sec.


Homework Equations


s=-1/2*a*t^2


The Attempt at a Solution


I find the acceleration in the to periods first

Acceleration = 19.4/10sec = 1.9 m/s^2
Deacceleration = -19.4/3sec. = -6-47 m/s^2


The I want to divide the travelled length for each individual timeperiod.

The first period (10 s) is = 1/2 * 1.9 * 10^2 = 95 meters
Seceond period (11 s) is = 19.4*11 = 213.4 meters
Third period (3 s) = -1/2*-6.47*3 = 9.7 meters

Travelled distance in all = 95+213.4+9.7 = 336.1 meters.

Is this correct?!




And yes I'm new here, so hello to everyone. Hope this can help me be better at phys/maths. I'm starting engineering soon, so I better get this handled!
 

Answers and Replies

  • #2
24
0
The complete equation is [tex]s=ut+\frac{1}{2}at^2[/tex].
For the first time period u=0 so your result looks ok, but for the third the initial velocity is u=19.4m/s which you havn't taken into consideration.
 
  • #3
6
0
Isn't that given in the deacceleration?

Deacceleration = -19.4/3sec. = -6-47 m/s^2 ?


Otherwise a hint would be cool!
 

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