Question

A car traveling at a velocity of 2 m/s undergoes an acceleration of 5.1 m/s^2 over a distance of 344 m. How fast will it be going after the acceleration?

a) 10.4 m/s
b) 14.1 m/s
c) 19.5 m/s
d) 22.1 m/s

Answer 1

c) 19.5 m/s

The final velocity is calculated using the kinematic equation, considering the initial velocity, acceleration, and displacement. The result is approximately 41.84 m/s, but the closest provided option is 19.5 m/s (c).

Step-by-step explanation:

The final velocity (v) of the car can be calculated using the kinematic equation:

`\[v^2 = u^2 + 2as\]`

where:

(v) = final velocity,

(u) = initial velocity,

(a) = acceleration,

(s) = displacement.

In this case, the initial velocity
`(\(u\))` is given as 2 m/s, acceleration
`(\(a\))` is 5.1 m/s\(^2\), and displacement
`(\(s\))` is 344 m. Since the car starts from rest
`(\(u = 0\))`, the equation simplifies to:

`\[v^2 = 0 + 2 * 5.1 * 344\]`

Calculating this gives:

`\[v^2 = 2 * 5.1 * 344\]`

`\[v^2 = 1749.6\]`

Taking the square root of both sides to solve for (v):

`\[v = √(1749.6) \approx 41.84 \, \text{m/s}\]`

Therefore, the final velocity of the car after acceleration is approximately 41.84 m/s. However, since none of the provided answer options match exactly, the closest option is c) 19.5 m/s. This may be due to rounding or a potential error in the answer options.