Application

• Oct 13th 2009, 09:01 AM
ur5pointos2slo
Application
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
• Oct 13th 2009, 09:31 AM
Defunkt
Quote:

Originally Posted by ur5pointos2slo
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

The question never mentioned the direction of the applied force in relation to the direction of the velocity... so I'll simply do the next steps in assumption that the force is in the same direction as the velocity.

(a) is correct, the rest are not. You should note that because the force applied on the automobile is not constant, neither is the acceleration, and thus $\displaystyle v=at$ is not correct.

For (b):

$\displaystyle \sum F = ma \Rightarrow a(t) = \frac{F}{m} = \frac{800-40t}{650}$
$\displaystyle v = \int a(t)dt = \int \frac{800-40t}{650}dt = \frac{800}{650}t - \frac{20}{650}t^2 + C$. Since $\displaystyle v_0 = 5ms^{-1}$, we get that $\displaystyle C=5$.
So we get: $\displaystyle v(t = 10) = \frac{80}{65}t - \frac{2}{65}t^2 + 5 = \frac{800}{65} - \frac{200}{65} + 5 = \frac{800-200}{65} + 5 = 5+\frac{600}{65} = 14.23ms^{-1}$

To solve (c), simply find $\displaystyle x(t) = \int v(t)dt$ and substitute t = 10 into the expression that you get.

(d) $\displaystyle v(t) = \frac{80}{65}t - \frac{2}{65}t^2 + 5 = 15$. Solve the quadratic with respect to t.
• Oct 13th 2009, 09:39 AM
ur5pointos2slo
I am not sure why but when I try to recalculate the velocity for myself I am getting 14.23 m/s.

(80/65)(10) - (2/65)*10^2 + 5 = 14.23?
• Oct 13th 2009, 09:48 AM
ur5pointos2slo
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• Oct 13th 2009, 09:56 AM
Defunkt
Quote:

Originally Posted by ur5pointos2slo
I am not sure why but when I try to recalculate the velocity for myself I am getting 14.23 m/s.

(80/65)(10) - (2/65)*10^2 + 5 = 14.23?

Yes, you're correct... that was stupid on my part. Fixed the original post now.
• Oct 13th 2009, 09:58 AM
ur5pointos2slo
Quote:

Originally Posted by Defunkt
Yes, you're correct... that was stupid on my part. Fixed the original post now.

Not a problem. I calculated it about 5 times over and once got the 3.154 so I must have typed it in wrong myself. Thanks for your help.