Cart is driven by a large propeller or fan, which can accelerate or decelerate the cart. The cart starts out at the position x=0m, with an initial velocity of +5.0ms and a constant acceleration due to the fan. The direction to the right is positive. The cart reaches a maximum position of x=+12.5m, where it begins to travel in the negative direction. Find the acceleration of the cart.

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Jeremy Merritt · Accepted Answer

Using the following equation of motion:  v2=v02+2ax  we can get the acceleration as:  a=v2−v022x (1)  The cart start moving with a speed of v0=5m⋅s−1 at x=0m, and reaches the maximum displacement at x=12.5m, at this petition the cart start to decelerating, so at this point the velocity is zero v=0, since the cart must come to a momentary halt before cart reverse its direction. Using (1), we can find the acceleration as:  a=0−(5m⋅s−1)22(12.5m)  =−1m⋅s−2  a=−1m⋅s−2

Bernard Lacey · Accepted Answer

Answer:  The acceleration of the cart is 1.0ms2 in the negative direction.  Explanation:  Using the equation of motion:  Vf2=Vi2+2⋅a⋅x  2⋅a⋅x=Vf2−Vi2  a=Vf2−Vi22⋅x  Where Vf is the final velocity of the cart, Vi is the initial velocity of the cart, a the acceleration of the cart and x the displacement of the cart.  Let x=Xf−Xi  Where Xf is the final position of the cart and Xi the initial position of the cart.  x=12.5−0  x=12.5  The cart comes to a stop before changing direction  Vf=0ms  a=02−522⋅12.5  a=−1ms2  The cart is decelerating  Therefore the acceleration of the cart is 1.0ms2 in the negative direction.

Nick Camelot · Accepted Answer

To solve the problem, we can use the kinematic equations of motion. The equation that relates the final position (x), initial velocity (v0), acceleration (a), and time (t) is given by:x=v0t+12at2Given that the cart starts at position x=0m with an initial velocity v0=+5.0m/s, we can substitute these values into the equation to find the time it takes for the cart to reach the maximum position:12.5=5.0t+12at2Now, we can differentiate this equation with respect to time to find the velocity equation:v=dxdt=5.0+atSince the cart reaches the maximum position at x=+12.5m, we can differentiate this equation to find the acceleration equation:a=dvdtLet&#039;s differentiate the velocity equation to find the acceleration expression:a=ddt(5.0+at)a=0+ddt(at)a=adtdta=aTherefore, the acceleration of the cart remains constant and does not depend on time. The acceleration of the cart is represented by the variable a.

Mr Solver · Accepted Answer

Answer:−25.0+50.0275.0−50.02m/s2Explanation:We can use the equation of motion to relate these quantities:xf=xi+vit+12at2Since the cart reaches its maximum position at xf=+12.5m, we can substitute the known values into the equation:+12.5=0+5.0t+12at2Simplifying the equation, we have:12at2+5.0t−12.5=0To solve this quadratic equation for a, we can use the quadratic formula:t=−b±b2−4ac2awhere a=12, b=5.0, and c=−12.5.Substituting the values into the quadratic formula, we have:t=−5.0±5.02−4·12·(−12.5)2·12Simplifying further:t=−5.0±25.0+25.01t=−5.0±50.01t=−5.0±521Since the cart starts at t=0 and reaches the maximum position at t=−5.0+521, we can substitute this value into the equation to solve for a:+12.5=0+5.0(−5.0+521)+12a(−5.0+521)2Simplifying the equation, we have:+12.5=0−25.0+25.02+12a(−5.0+521)2+12.5=−25.0+25.02+12a(−5.0+52)2Now we can solve for a by rearranging the equation:12a(−5.0+52)2=+12.5−25.0+25.0212a(−5.0+52)2=−12.5+25.02a(−5.0+52)2=2(−12.5+25.02)a=2(−12.5+25.02)(−5.0+52)2Simplifying further:a=−25.0+50.02(−5.0+52)(−5.0+52)a=−25.0+50.0225.0−50.02+25.0·2a=−25.0+50.0275.0−50.02Therefore, the acceleration of the cart is −25.0+50.0275.0−50.02m/s2.

madeleinejames20 · Accepted Answer

Solution:Using the equation of motion:x=x0+v0t+12at2where t is the time taken.We are given the following information:x0=0m (initial position)v0=+5.0m/s (initial velocity)x=+12.5m (final position)From the given information, we know that the cart travels from x0 to x and changes direction at x=12.5m.To find the acceleration a, we can rearrange the equation of motion as follows:a=2(x−x0−v0t)t2Substituting the known values:a=2(12.5m−0m−(+5.0m/s)t)t2Simplifying the expression:a=25m−10m/s·tt2Therefore, the acceleration of the cart is a=25m−10m/s·tt2.

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