# Distance traveled in nth second

## Trending Questions

**Q.**A small block slides down on a smooth inclined plane, starting from rest at time t=0. Let Sn be the distance travelled by the block in the interval, t=n−1 to t=n, then, the ratio SnSn+1 is

- 2n2n−1
- 2n−12n
- 2n−12n+1
- 2n+12n−1

**Q.**

The initial velocity of a body moving along a straight line is$7m/s$. It has a uniform acceleration of$4m/{s}^{2}$ The distance covered by the body in the$5th$ second of its motion is

$25m$

$35m$

$50m$

$85m$

**Q.**A small block slides down on a smooth inclined plane, starting from rest at time t=0. Let Sn be the distance travelled by the block in the interval, t=n−1 to t=n, then, the ratio SnSn+1 is

- 2n2n−1
- 2n−12n
- 2n−12n+1
- 2n+12n−1

**Q.**If a body starts from rest and travels 120 cm in the 6th second, then the acceleration of the body is

- 0.2 m/s2
- 0.027 m/s2
- 0.218 m/s2
- 0.03 m/s2

**Q.**The initial velocity of a body moving along a straight line is 7 ms. It has a uniform acceleration of 4 ms2. The distance covered by the body in the 5th second of its motion is

- 25m
- 35m
- 50m
- 85m

**Q.**A particle experiences a constant acceleration for 20 s starting from rest. If it travels a distance S1 in the first 10 s and a distance S2 in the next 10 s, then:

- S1=S2
- S1=S23
- S1=S22
- S1=S24

**Q.**The ratio of distances covered by a freely falling body (starting from rest) during the 1st, 2nd, 3rd second of its motion is: [Take g=10 m/s2]

- 3:6:9
- 1:3:5
- 2:3:5
- 1:3:7

**Q.**A body covers 26 and 28 meters in 10th and 11th seconds respectively. The body starts

- from rest and moves with uniform velocity.
- from rest and moves with uniform acceleration.
- with an initial velocity and moves with uniform acceleration.
- with an initial velocity and moves with uniform velocity.

**Q.**The initial velocity of a body moving along a straight line is 7 ms. It has a uniform acceleration of 4 ms2. The distance covered by the body in the 5th second of its motion is

- 25m
- 35m
- 50m
- 85m

**Q.**The initial velocity of a particle is 10 m/s and its retardation is 2 m/s2. The distance moved by the particle in 5th second of its motion is:

- 1 m
- 19 m
- 50 m
- 75 m

**Q.**A body A starts from rest with an acceleration a1. After 2 seconds, another body B starts from rest with an acceleration a2. If they travel equal distances in the 5th second, after the start of A, then the ratio a1:a2 is equal to

- 5 : 9
- 5 : 7
- 9 : 5
- 9 : 7

**Q.**A particle experiences a constant acceleration for 20 s starting from rest. If it travels a distance S1 in the first 10 s and a distance S2 in the next 10 s, then:

- S1=S2
- S1=S23
- S1=S22
- S1=S24

**Q.**A particle is dropped from the tower. The distance travelled by it in the last second is 15 m. The height of the tower is

[Take g=10 m/s2]

- 22.5 m
- 20 m
- 25 m
- 30 m

**Q.**A body is freely falling under the action of gravity. It covers half the total distance in the last second of its fall. If it falls for n second, then the value of n is

- √2
- 2+√2
- 2
- 2−√2

**Q.**A body moving in a straight line with constant acceleration of 10 ms−2 covers a distance of 40 m in the 4th second. How much distance will it cover in the 6th second?

- 50 m
- 60 m
- 70 m
- 80 m

**Q.**A space shuttle moving at 2 m/s at t=0 begins to accelerate at a constant rate of 2 m/s2. If Sn is the distance covered by it in the nth second, then:

- S1+S4>S2+S3
- S1+S4=S2+S3
- S1+S4<S2+S3

**Q.**A car starts accelerating with an acceleration of 13 m/s2 when its speed was 18 km/h. If the displacement of car in nth minute after it starts accelerating is 4.8 km, then n is:

**Q.**A body A starts from rest with an acceleration aA. After 2 seconds, another body B starts from rest with an acceleration aB. If they travel equal distances in the 5th second, after the start of A, then the ratio aA:aB is equal to

- 5:9
- 5:7
- 9:5
- 9:7