Question

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 × 102 meters

Answer 1

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.

$6.96 \times 102$ meters

B.

$1.27 \times 103$ meters

C.

$5.70 \times 102$ meters

D.

$1.26 \times 102$ meters

E.

$6.28 \times 102$ meters

Solution:

$V_{1} = 58\frac{\mathrm{m}}{\mathrm{s}} -$ the initial velocity of the spaceship;

$V_{2} = 153\frac{\mathrm{m}}{\mathrm{s}} -$ final velocity of the spaceship;

d - distance that spaceship covered after 12 s;

$t = 12s -$ time to cover the distance d;

Assuming constant acceleration we can use the rate equation and motion equation the to find the the distance that spacesip covered after 12 s. Rate equation alond the X axis:

$V_{2} = V_{1} + at$

$$a = \frac {V _ {2} - V _ {1}}{t}$$

Motion equation alond the Y axis:

$$\mathrm {d} = \mathrm {V} _ {1} \mathrm {t} + \frac {\mathrm {a t} ^ {2}}{2}$$

(1)in(2):

$$\begin{array}{l} \mathrm{d} = \mathrm{V}_{1} \mathrm{t} + \frac{\left(\frac{\mathrm{V}_{2} - \mathrm{V}_{1}}{\mathrm{t}}\right) \mathrm{t}^{2}}{2} = \mathrm{V}_{1} \mathrm{t} + \frac{(\mathrm{V}_{2} - \mathrm{V}_{1}) \mathrm{t}}{2} = \frac{(\mathrm{V}_{1} + \mathrm{V}_{2}) \mathrm{t}}{2} = \frac{\left(153 \frac{\mathrm{m}}{\mathrm{s}} + 58 \frac{\mathrm{m}}{\mathrm{s}}\right) \cdot 12 \mathrm{s}}{2} = \\ = 12.7 \times 10^{3} \mathrm{m} \end{array}$$

Answer: distance that spaceship covered after 12s is B) $12.7 \times 10^{3} \mathrm{~m}$