Question

A spaceship is moving in space with a velocity of 60 kms⁻¹. It fires its retro engines for 20 seconds and velocity is reduced to 55 kms⁻¹. Calculate the distance travelled by a spaceship in 40 s, from the time of firing of the retro- rockets.

Answer 1

Initial velocity of spaceship = u = 60 kms⁻¹

Final velocity of spaceship = v = 55 kms⁻¹

It decelerates for 20 s

t = 20 s

v = u + at

55 = 60 + a (20)

20a = 55 – 60

20a = −5

a = `(-5)/20`

a = − 0.25

distance travelled in the first 20 sec. = s = u × t − `1/2` a × t²

s = 60 × 20 − `1/2` × 0.25 × 20²

= 1200 − `1/2` × 0.25 × 400

= 1200 − `1/2` × 100

= 1200 − 50

= 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. = 1,150 + 1,100 = 2,250 km