Not sure where to start for number 4 and 5. I think I got the first part of 4 right but unsure!! Any help is greatly appreciated!!!

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Not sure where to start for number 4 and 5. I think I got the first part of 4 right but unsure!! Any help is greatly appreciated!!!

Physics
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5. A 39,250 N force is applied to a 7200kg car initially at rest . what is the acceleration rate? How fast will the car be moving after 3,8 seconds? Sorry if you couldn't see the picture.
use newton's second law \(F=ma\)
39,250 =7200 *a 39,250/7200 =a=5.5 m/s

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careful about the units a=5.5 m/s^2
ok. How do I calculate how fast it's going to be after 3.8 seconds. Do I use v=10 * t but that has something to od with free fall? So, no?
so the car is being accelerated for 3.8 seconds
\(a = 5.5\) \(t=3.8\) initial velocity, \(u=0\) final velocity, \(v=?\)
5.58 * 3.8 =20.9 m/s
right, but it seems you have a rounding error.. maybe just enter below into the calculator : 39250/7200*3.8 = 20.7 m/s
ok . I think I understand. Thank you so much!!!!
yw!
as you can see, "change in speed" requires acceleration no acceleration means no "change in speed"

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