Calculate the acceleration of a 1400-kg car that stops from 32 km/h "on a dime" (on a distance of 1.7 cm).

-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 = -  

=> 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.

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures

+15-35=-20

going in opposite directions