Velocity and acceleration problem

[-Physician]

Homework Statement

The car goes with a initial velocity of 234km/h, and accelerates 9m/s² for time of 20s. Find the displacement of the car in accelerating uniformly motion and the final velocity. Also the find the change in velocity !$Δv$

Homework Equations

$v_i=234\frac{km}{h}=234\frac{1m}{3.6s}$
$a=9\frac{m}{s^2}$
$t=20s$
$d=v_0t+\frac{1}{2}at^2$(Find)
$v_f=v_0+at$(Find)
$Δv=v_f-v_i$ (Find)

The Attempt at a Solution

Just waiting an answer :) thanks! (i know you may not answer because you don't do our homework but you help us, but i really need help for this)

 

[LawrenceC]

You have the equations to use. The speed is given in km/hr but the acceleration rate is in m/sec^2. You must convert to like units before adding the terms in the displacement equation.

 

[-Physician]

I see, but

accelerating uniformly motion

, what does that mean, does anything change?

 

[LawrenceC]

It should be stated "accelerating uniform motion". Motion is a noun so it needs an adjective not an adverb as a modifier.

That said, the term means that the acceleration is constant. The equation is not valid unless the acceleration is constant over the specified time, t.

 

[-Physician]

Oh thanks so:
$v=234 \frac{km}{h}=234 \frac{1m}{3.6s}=65 \frac{m}{s}$
$v_f=v_i+at=65 \frac{m}{s} + (9 \frac{m}{s^2})(20s)=65 \frac{m}{s} + 180 \frac{m}{s}=245 \frac{m}{s}$
$d=v_0 t + \frac{1}{2} at^2=(65 \frac{m}{s})(20s)+ \frac{1}{2}(9 \frac{m}{s^2})(20s)^2=1300m+1800m=3100m=3.1km$
Is that correct?

 

[LawrenceC]

Yes, it is correct.