What is the Velocity and Distance of an Automobile with a Changing Force?

[ur5pointos2sl]

I have posted a problem below and my attempt could someone please tell me if I am going wrong and where. Thanks

An automobile of mass 650 kg is acted on by a net force (F) given by F = 800 - 40.0t, where t is in seconds and F is in Newtons. At t=0, the velocity of the automobile is 5.00m/s.

a) Find the acceleration of the automobile at t=10.0s

800-40(t) / 650 = 400/650 = .615 m/s^2

b) Find the velocity of the automobile at t=10.0s

.615t = .615(10s) =6.15 m/s

c) Find the distance covered in this time.

x = (1/2)V*t
(1/2)(6.15)(10) = 30.75 m

d) At what time will the automobile be moving with a velocity of 15.0 m/s

V=at

t=V/a = 15/.615
t= 24 s

 

[Suy]

b is incorrect
remember a=(final velocity-initial velocity)/ (delta time)

 

[ur5pointos2sl]

b is incorrect
remember a=(final velocity-initial velocity)/ (delta time)

In this problem there is no final velocity?

 

[Suy]

At t=0, the velocity of the automobile is 5.00m/s.
For B, you are solving the final velocity at 10.0s, (initial is not 0)

 

[ur5pointos2sl]

At t=0, the velocity of the automobile is 5.00m/s.
For B, you are solving the final velocity at 10.0s, (initial is not 0)

ok correct me if I am wrong..I am a beginner at this..

So if I rearrange the equation you gave me

final velocity = initial velocity + acceleration * time
= 5.00 + (.615 * 10.0 )
= 11.15 m/s ?

 

[Suy]

Correct