Question

select the correct answer.
a spaceship moving with an initial velocity of 58.0 meters/second experiences a uniform acceleration and attains a final velocity of 153 meters/second. what distance has the spaceship covered after 12.0 seconds?
a. 6.96 × 10² meters
b. 1.27 × 10³ meters
c. 5.70 × 10² meters
d. 1.26 × 10² meters
e. 6.28 × 10⁷ meters

Explanation:

Step1: Recall kinematic distance formula

The formula for distance covered with uniform acceleration when initial velocity ($v_0$), final velocity ($v$), and time ($t$) are known is:
$$d = \frac{v_0 + v}{2} \times t$$

Step2: Substitute given values

$v_0 = 58.0 \, \text{m/s}$, $v = 153 \, \text{m/s}$, $t = 12.0 \, \text{s}$
$$d = \frac{58.0 + 153}{2} \times 12.0$$

Step3: Calculate the average velocity

$$\frac{58.0 + 153}{2} = \frac{211}{2} = 105.5 \, \text{m/s}$$

Step4: Compute total distance

$$d = 105.5 \times 12.0 = 1266 \, \text{meters} = 1.27 \times 10^3 \, \text{meters}$$

Answer:

B. $1.27 \times 10^3$ meters