Motion in a Straight Line
- The distance travelled by a particle starting from rest and moving with an acceleration 4/3 ms–2, in the third second is:
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Distance travelled in the nth second is given by tn = u + a (2n - 1) 2
put u = 0, 4/3 ms-2 , n = 3∴ d = 0 + 4 (2 × 3 –1) = 4 × 5 = 10 m 3 × 2 6 3 Correct Option: C
Distance travelled in the nth second is given by tn = u + a (2n - 1) 2
put u = 0, 4/3 ms-2 , n = 3∴ d = 0 + 4 (2 × 3 –1) = 4 × 5 = 10 m 3 × 2 6 3
- What will be the ratio of the distances moved by a freely falling body from rest in 4th and 5th seconds of journey?
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x(4) = g/2 (2 × 4 - 1) = 7 x(5) g/2 (2 × 5 - 1) 9
[∵ Sn th = u + a/2 (2n - 1) and u = 0, a = g]Correct Option: B
x(4) = g/2 (2 × 4 - 1) = 7 x(5) g/2 (2 × 5 - 1) 9
[∵ Sn th = u + a/2 (2n - 1) and u = 0, a = g]
- The position x of a particle with respect to time t along x-axis is given by x = 9t² – t³ where x is in metres and t in second. What will be the position of this particle when it achieves maximum speed along the +ve x direction?
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Speed v = dx = d (9t² - t³) = 18t – 3t² dt dt
Correct Option: A
Speed v = dx = d (9t² - t³) = 18t – 3t² dt dt
- A particle moving along x-axis has acceleration f, at time t, given by f = f0(1 - (t/T)) , where f0 and T are constants. The particle at t = 0 has zero velocity. In the time interval between t = 0 and the instant when f = 0, the particle’s velocity (vx) is
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Here, ƒ = ƒ0 
1 - t 
or, dv = ƒ0 1 - t T dt T or, dv = ƒ0 1 - t dt T ∴ v = ∫ dv = ∫ 
ƒ0 1 - t 
dt T or v = ƒ0 t - t² + C 2T
where C is the constant of integration.
At t = 0, v = 0.∴ 0 = ƒ0 0 - 0 + C ⇒ C = 0 2T ∴ v = ƒ0 t - t² 2T
if ƒ = 0, then0 = ƒ0 t - t² ⇒ t = T 2T
Hence, particle's velocity in the time interval t = 0 and t = T is given byvx = t = T∫t = 0 dv = T∫t = 0 ƒ0 1 - t dt T = ƒ0 t - t² T 2T 0 = ƒ0 T - T² = ƒ0 T - T = 1 ƒ0T 2T 2 2 Correct Option: C
Here, ƒ = ƒ0 1 - t or, dv = ƒ0 1 - t T dt T or, dv = ƒ0 1 - t dt T ∴ v = ∫ dv = ∫ ƒ0 1 - t dt T or v = ƒ0 t - t² + C 2T
where C is the constant of integration.
At t = 0, v = 0.∴ 0 = ƒ0 0 - 0 + C ⇒ C = 0 2T ∴ v = ƒ0 t - t² 2T
if ƒ = 0, then0 = ƒ0 t - t² ⇒ t = T 2T
Hence, particle's velocity in the time interval t = 0 and t = T is given byvx = t = T∫t = 0 dv = T∫t = 0 ƒ0 1 - t dt T = ƒ0 t - t² T 2T 0 = ƒ0 T - T² = ƒ0 T - T = 1 ƒ0T 2T 2 2
- A particle moves along a straight line OX. At a time t (in seconds) the distance x (in metres) of the particle from O is given by x = 40 + 12t – t³. How long would the particle travel before coming to rest?
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x = 40 + 12 t – t³
For v = 0; t = √12/3 = 2sec
So, after 2 seconds velocity becomes zero.
Value of x in 2 secs = 40 + 12 × 2 – 2³
= 40 + 24 – 8 = 56 mCorrect Option: B
x = 40 + 12 t – t³
For v = 0; t = √12/3 = 2sec
So, after 2 seconds velocity becomes zero.
Value of x in 2 secs = 40 + 12 × 2 – 2³
= 40 + 24 – 8 = 56 m
