Question

A car has an initial velocity of 15m/s and an acceleration of 1m/(s^2). How far does the car go in the first 10s after the acceleration begins?

Answer 1

To find the distance the car travels during the first 10 seconds after the acceleration begins, we can use the kinematic equation:

$d=v_{i}t+\frac{1}{2}a{t}^2$

where: $d$ is the distance traveled, $v_i$ is the initial velocity, $a$ is the acceleration, and $t$ is the time.

Given: $v_i=15\text{m/s}$, $a=1{\text{m/s}}^2$, and $t=10\text{s}$.

Substituting these values into the equation, we have:

d=(15 m/s)(10 s)+12(1 m/s2)(10 s)2d = (15 , ext{m/s})(10 , ext{s}) + frac{1}{2}(1 , ext{m/s}^2)(10 , ext{s})^2d=(15m/s)(10s)+21</s