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

User Ariel
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Final answer:

The acceleration of a 1400-kg car that goes from 32 km/h to 0 km/h over a distance of 1.7 cm is approximately -2328.68 m/s². This calculation involves converting units and applying a kinematic equation.

Step-by-step explanation:

To calculate the acceleration of a 1400-kg car that stops from 32 km/h to 0 km/h over a distance of 1.7 cm, we first need to convert units and apply the kinematic equation. The initial velocity (v0) must be converted from kilometers per hour to meters per second:

v0 = 32 km/h * (1000 m/km) * (1 h/3600 s) = 8.89 m/s

The final velocity (v) is 0 m/s since the car stops, and the displacement (s) is 1.7 cm or 0.017 m. We can use the kinematic equation:

v2 = v02 + 2as

Plugging in the known values:

0 = (8.89 m/s)2 + 2 * a * 0.017 m

Solving for acceleration (a), we get:

a = -(8.89 m/s)2 / (2 * 0.017 m)

This gives us an acceleration of approximately -2328.68 m/s2, which is a very high deceleration and likely not achievable under normal circumstances, indicating a highly idealized or theoretical scenario.

User Mervasdayi
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