Question

a spaceship is moving in a space with a velocity of 60 km per second it 5800 engines for 22nd and velocity is reduced to 55 km per second calculate the distance travelled by the spaceship in 40 seconds from the time of filing of retro rocket

Answer 1

Hey

$given$
U = 60 km/sec
v = 55 km/sec
t = 20 sec
let us assume that a uniform deceleration is resulted by firing the retro engines.
   a = (v - u)/ t = - 5/20 = - 0.25 km/s²
distance travelled in the first 20 sec. = s = u t + 1/2 a t²
   s = 60 * 20 - 1/2 * 0.25 * 20² = 1,150 km.

After 20 sec. the velocity of the spaceship is constant at 55 km/sec.  Hence, the distance travelled is = 55 * 20 = 1, 100 km

The total distance travelled in 40 sec. = 2, 250 km

Answer 2

hope this answer helps you..

please mark it as brainliest answer..