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

The spaceship moving with uniform acceleration the distance covered in t s= 1/2(u+v) times t = 1/2(58+153) times 12
=1266m

Answer 2

Answer:

b) 1.27 × 103 meters

Explanation:

For a uniform acceleration, given an initial, a final velocity, and time, the distance traveled can be calculated with the formula:

$x=\frac{V_{1}+V_{2} }{2}t$

Plugging in the given values in the equation:

$x=\frac{58m/s+153m/s }{2}*12s=1266m=1.266*10^3m=1.27*10^3m$

So, the distance the spaceship covered after 12.0 seconds is 1.27*10^3 meters.