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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.
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Solution
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
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