A car moves from the position A to B along the circular road AB (rho =500ft) fro

Question

A car moves from the position A to B along the circular road AB (rho =500ft) from rest. While the car is moving along this circular road AB, the car increases its speed with a constant rate. When the car arrives at the point B, the car maintains its acceleration so that it can pass through the point C with the speed of 100 ft/s. On the other side of the road, a motorcycle also moves from the point D to E with a constant negative acceleration of -5 ft/s^2 (i.e. decelerating). Determine the tangential and normal components of the car's acceleration at time t=14s Treating both bodies as particles, find the car's acceleration relative to the motorcycle at time t=14s (either as a Cartesian vector or as a magnitude AND direction)

Explanation / Answer

Hi,

As I like to work in the international system of units I will transform all the data:

vc = 100 ft/s = 30.48 m/s ; R = 500 ft = 152.4 m ; a2 = -5 ft/s2 = -1.524 m/s2 ; d = 150 ft = 45.72 m

1. We need to find the value of the tangential acceleration (at ) because it remains constant through the motion of the car and it is useful to find other magnitudes.

We know the speed of the car at point C, and we know the distance travelled by the car from A to C, and we also know that the car begins its movement from rest so:

vc2 = 2at (S + d) ; where S is the distance travelled along the circular road (from A to B), which can be calculated as:

S = R ; where the angle must be in radians.

In this case, the angle should be 90° (/2) according to the figure, so:

S = (/2)(152.4 m) = 239.389 m

With this value we can find the tangential acceleration:

at= vc2 /[ 2(S + d) ] = (30.48 m/s)2 /[ 2(239.389 m + 45.72 m] = 1.629 m/s2

We need to know if t = 14 s occurs when the car is still in the circular road or it is after that. In order to know we can apply the following equation:

S = (1/2) att2 ; which can be applied if the motion stars from rest and the acceleration is constant

t = [ 2S/at ]1/2 = [ 2*239.389 m/ 1.629 m/s2 ]1/2 = 17.144 s

As this time is bigger than 14 s, and this is the time that the car takes to move from A to B, we have to conclude that the time given is within the distance AB.

If the previous is true, then the normal component of the acceleration (which is the centripetal acceleration) is:

v = at t = (1.629 m/s2) (14 s) = 22.806 m/s

ac = v2 / R = ( 22.806 m/s )2 /(152.4 m) = 3.413 m/s2

So, the acceleration at 14 s is:

a = [ -3.413 ; 1.629 ] m/s2 = [ -11.2 ; 5.4 ] ft/s2

Note: the negative sign in normal component is because this one is directed towards the center of the circle.

2. For this part I think is better to use a cartesian reference to write the vector. We take the positive x axis going to the right, and the positive y axis going upwards.

At 14 s, the car has travelled:

S' = (1/2) at t2 = (1/2) (1.629 m/s2) (14 s)2 = 159.64 m

So, the angle between the normal component of the acceleration and the horizontal is:

= S'/R = 1.048 rad = 60.02 ° (below the horizontal)

Therefore, the angle between the tangential component of the acceleration and the horizontal is:

= 90° - = 29.98° (above the horizontal)

the components of the acceleration expressed in this new system are:

atx = at cos(29.98°) = 1.452 m/s2 ;    aty = at sin(29.98°) = 0.739 m/s2

anx = an cos(-60.02°) = 2.005 m/s2  ; any = an sin(-60.02°) = -2.762 m/s2

The acceleration vector would be:

ax = 3.457 m/s2   ; ay = -2.023 m/s2

Using the transformation equations we can find the acceleration of the car as it is perceived by the motorcycle:

amc = am + ac

The components of the acceleration of the motorcycle (am ) are:

amx = am cos(210°) = -1.505 m/s2   ; amy= am sin(210°) = -0.238 m/s2

So, the vector of the acceleration of the car at 14 s from the perspective of the motorcycle would be:

ax = 1.952 m/s2 = 6.4 ft/s2   ; ay = -2.261 m/s2= -7.4 ft/s2

I hope it helps.

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