Question

A dragster, starting from rest, accelerates at 28 m/s/s for 4.5 s. Then the parachute comes out and it slows to 46 m/s in 10 s. What is its acceleration in those 10 s when the parachute is deployed

Answer 1

Let's first find the velocity of the dragster just before the parachute comes out:

$$v=v_0+at,$$

here, $v$ is the velocity of the dragster just before the parachute comes out, $v_0=0\ \dfrac{m}{s}$ is the initial velocity of the dragster, $a=28\ \dfrac{m}{s^2}$ is the acceleration of the dragster during time $t=4.5\ s.$

Then, from this equation we can find $v$:

$$v=at=28\ \dfrac{m}{s^2}\cdot 4.5\ s=126\ \dfrac{m}{s}.$$

Finally, we can find the acceleration of the draster in those 10 s when the parachute is deployed from the same kinematic equation:

$$v=v_0+at,$$

here, $v=46\ \dfrac{m}{s}$ is the final velocity of the dragster when the parachute is deployed, $v_0=126\ \dfrac{m}{s}$ is the initial velocity of the dragster just before the parachute comes out, $t=10\ s$ is the time during which the dragster slows down.

Then, from this equation we can find the acceleration of the draster in those 10 s when the parachute is deployed:

$$a=\dfrac{v-v_0}{t}=\dfrac{46\ \dfrac{m}{s}-126\ \dfrac{m}{s}}{10\ s}=-8\ \dfrac{m}{s^2}.$$

The sign minus indicates that the dragster decelerates.

Answer:

$a=-8\ \dfrac{m}{s^2}.$