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a car accelerates uniformly from 17.8 m/s^2 to 40.7 m/s^2 in 7.5 second…

Question

a car accelerates uniformly from 17.8 m/s^2 to 40.7 m/s^2 in 7.5 seconds. what was the acceleration of the car? 2.8 m/s^2 5.92 m/s^2 3.05 m/s^2 6.09 m/s^2

Explanation:

Step1: Identify acceleration formula

The formula for acceleration is $a=\frac{v - u}{t}$, where $v$ is final - velocity, $u$ is initial - velocity, and $t$ is time.

Step2: Substitute values

Given $u = 17.8\ m/s$, $v = 40.7\ m/s$, and $t = 7.5\ s$. Then $a=\frac{40.7 - 17.8}{7.5}$.

Step3: Calculate numerator

$40.7-17.8 = 22.9$.

Step4: Calculate acceleration

$a=\frac{22.9}{7.5}\approx3.05\ m/s^{2}$.

Answer:

$3.05\ m/s^{2}$