a car turns into a driveway that slopes upward at a 9degree angle, car is moving at 6.5m/s. if the driver lets the car coast, how far along the slope will the car roll before being instantaneously at rest and start rolling back? please show kinematic steps.

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a car turns into a driveway that slopes upward at a 9degree angle, car is moving at 6.5m/s. if the driver lets the car coast, how far along the slope will the car roll before being instantaneously at rest and start rolling back? please show kinematic steps.

Physics
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? sry I cant help you I honestly have no clue
|dw:1441424417873:dw|
okay so there is an acceleration on this far in the opposite direction parallel to the ramp like so |dw:1441424501231:dw|

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|dw:1441424541598:dw|
can you figure out what that acceleration must be now
Hi, are you here?
we can represent your exercise as below: |dw:1441470574464:dw|
the equations of the motion of the car, provided that its position at initial time t=0, is the origin of my coordinate system, are: \[\Large \begin{gathered} x = - \frac{1}{2}g\left( {\sin \theta } \right){t^2} + {v_0}t \hfill \\ \hfill \\ v = - g\left( {\sin \theta } \right)t + {v_0} \hfill \\ \end{gathered} \] where \( \large \theta = 9 \) degrees, \( \Large v_0= 6.5 m/sec \) is the initial speed. Please note that on your car there is an acceleration \( \Large -g \sin \theta \) At the time \( \Large \tau \) when the car will reach the rest position, its speed is equal to zero, so we can write: \[\Large 0 = - g\left( {\sin \theta } \right)\tau + {v_0}\quad \Rightarrow \tau = \frac{{{v_0}}}{{g\sin \theta }}\] so, substituting into the first equation, we wil get the requested distance: \[\Large \begin{gathered} d = - \frac{1}{2}g\left( {\sin \theta } \right){\left( {\frac{{{v_0}}}{{g\sin \theta }}} \right)^2} + {v_0}\frac{{{v_0}}}{{g\sin \theta }} = \hfill \\ \hfill \\ = \frac{{v_0^2}}{{2g\sin \theta }} \hfill \\ \end{gathered} \]
oh i just realized we can also just use kinetic and potential energy k1+p1=k2+p2, only kinetic in beginning, only potential at end 1/2mv^2=mgh 1/2v^2=gh h=(v^2/2g) h/sin(theta)=d |dw:1441515361286:dw| = distance along the track
|dw:1441534416544:dw| Clearly the car has an acceleration in parallel to the slope, opposite to the car's motion Let's give sign to the motion opposite to the car as negative and in direction of car as positive so acceleration of the car is equal to the force acting in the direction opposite to the motion of the car divided by the mass of car \[a=\frac{F}{m}=-\frac{mgsin(\theta)}{m}=-g\sin(\theta) ms^{-2}\] The negative sign as stated implies retardation or deacceleration of the car Also given initial velocity of car as \[u=+6.5ms^{-1}\] Now we are considering the motion of the car from the start to the point where it is at rest so the final velocity (v) will be the car's velocity at rest, a body at rest has 0 velocity so we have final velocity \[v=0ms^{-1}\] Now we are required to find the displacement of the car from the initial position, apply kinematic equation for uniform accelerated motion we have \[v^{2}-u^{2}=2as\] Since v is 0 \[-u^{2}=2as\] or \[s=\frac{-u^2}{2a}\]\[s=\frac{-u^2}{-2g\sin(\theta)}m\] \[s=\frac{u^2}{2g\sin(\theta)}m\] You already given the angle theta, the initial velocity and value of g is constant Whenever value of g is not specified, you must use the value \[g=9.8ms^{-2}\]
thank you so much for your help guys, I got it now.

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