Question

A 2.9×103 kg car accelerates from rest under
the action of two forces. One is a forward
force of 1153 N provided by traction between
the wheels and the road. The other is a 920 N
resistive force due to various frictional forces.
How far must the car travel for its speed to
reach 2.3 m/s?
Answer in units of m

Answer 1

If we define the x direction as positive in the same direction as the accelleration
\[\sum_{}^{}Fx = 1153N - 920N = M*a\]
\[233N = (2.9*103kg)*a\]
\[233N = (298.7kg)*a\]
\[233N/298.7kg = a\]
\[a \approx 0.78005(m/s) \]
Using the equation: \[v = vi + a*t\]
\[2.3(m/s) = 0.78005(m/s^2)t\]
\[t \approx 2.94853(s)\]
Using the equation: \[d = vi*t + (t*a^2)/2\]
\[d \approx 1.79412(m)\]
Therefore, the car must travel approximately 1.79 meters to reach a speed of 2.3m/s

Answer 2

I tried that answer and it didn't work :/

Answer 3

I think jziggy is right. but I think there was a mix up from the post to the solution.
the phrase:
"A 2.9×103 kg car"
should have been
"A 2.9×10^3 kg car"

I worked it out and it turns out:
A 2.9×103 kg car = 658 Lb car [which is lighter than a motorcycle]
A 2.9×10^3 kg car = 6,393 Lb car [which is more believable]

So, from the second line of the equations posted, all the way to the end can be re-worked in the same way using the car mass of 2900 kg to get the right answer. ^_^

Answer 4

I tried to rework the problem but I think I may have done something wrong O.o

Answer 5

what do you get for a? and for t?

Answer 6

@RavenLynette

Answer 7

for t I got 28.6265 and for a I got .080345 :/

Answer 8

l or t is 0.080345

Answer 9

Maybe that's right

Answer 10

time is that?

Answer 11

a is right 0.0803

Answer 12

t = 28.64 is correct

Answer 13

I got .09 something for the distance

Answer 14

distance should be = .5*.0803*28.6^2 = 28.67 meters
\[d=vi∗t+(1/2)*a*t^{2}\]

Answer 15

@RavenLynette you 'see' it?

Answer 16

Oh ok thanks :)

Answer 17

^_^