Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
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
APPEARS IN
संबंधित प्रश्न
What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?
What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?
A driver of a car travelling at 52 km h−1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km h−1 in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?
What can you say about the motion of a body whose distance-time graph is a straight line parallel to the time axis ?
Fill in the following blank with suitable word :
The slope of a distance-time graph indicates ………………….. of a moving body.
Name the two quantities, the slope of whose graph give speed .
A cyclist is travelling at 15 m s-1. She applies brakes so that she does not collide with a wall 18 m away. What deceleration must she have ?
Two friends leave Delhi for Chandigarh in their cars. A starts at 5 am and moves with a constant speed of 30 km/h, whereas B starts at 6 am and moves with a constant speed of 40 kmh-1. Plot the distance-time graph for their motion and find at what time the two friends will meet and at what distance from Delhi.
Two students were asked to plot a distance-time graph for the motion described by Table A and Table B.
| Table A |
||||||
| Distance moved (m) | 0 | 10 | 20 | 30 | 40 | 50 |
| Time (minutes) | 0 | 2 | 4 | 6 | 8 | 10 |
| Table B |
||||||
| Distance moved (m) | 0 | 5 | 10 | 15 | 20 | 25 |
| Time (minutes) | 0 | 1 | 2 | 3 | 4 | 5 |
The graph given in figure is true for
The slope of the distance-time graph indicates the speed.
