-2324.4m/s²
Use one of the equations of motion as follows;
v² = u² + 2as ------------------------(i)
Where;
v = final velocity of object
u = initial velocity of object
a = acceleration of the object
s = distance covered by the object
From the question, the object is the car and it has the following;
v = final velocity of the car = 0 (since it comes to a stop)
u = initial velocity of the car = 32km/h = (32 x 1000/3600) m/s = 8.89m/s
a = unknown
s = on a dime = 1.7cm = 0.017m
Substitute these values into equation (i) as follows;
=> 0² = 8.89² + (2 x a x 0.017)
=> 0 = 79.03 + (0.034a)
=> 0.034a = 0 - 79.03
=> 0.034a = - 79.03
=> a = - [tex]\frac{79.03}{0.034}[/tex]
=> a = - 2324.4m/s²
Therefore, the acceleration of the car is -2324.4m/s².
Note that the negative sign indicates that the car is actually decelerating.
