If a particle moves in a straight line, it is called a rectilinear motion. Suppose if it moves with uniform acceleration in straight line, it can call as uniform rectilinear motion (accelerated uniformly). Based on the above assumption, we have the following formulae to deal with:
(i) v = u + at
(ii) s = (1/2) ( u +v) = ut + (1/2) at2.
(iii) v2 = u2 + 2as.
Here s is the displacement, u the initial velocity and v is the final velocity, t the time taken, a is the acceleration ( if it is positive) or retardation( if it is negative).
Ex 1: If a car reaches a velocity of 72 km/hr in 10s which h has started from rest , calculate:
(i) the acceleration
(ii) the average velocity.
Sol: Given: u = 0 ( rest )
v = 72 km/hr
= `(72 xx1000)/(60xx60)`
= 20 m/s
t = 10s.
(i) Acceleration, a = `(v** u )/t`
= `(20** 0)/10`
= 2 m/s2.
(ii) Average velocity = `(20 +0)/2` = 10m/s.
Ex 2: A mini van initially at rest starts moving with the uniform acceleration of 0.5 m/s2 and travels a distance 25m. Find:
(i) its final velocity,
(ii) the time taken.
Sol: Given: Initial velocity, u = 0,
Acceleration, a = 0.5 m/s2
Distance, s = 25m.
(i) From, v2 = u2 +2as
We have, v2 = 02 +2(0.5)(25)
v = 5 m/s.
(ii) From, v = u + at
T = `( v ** u)/a`
= `(5 -0)/0.5`
Ex 3: A ball starts move in straight line from a point with velocity 10 m/s and acceleration -2 m/s2, find the position and the velocity of the particle at (i) t = 5s (ii) t = 10s.
Sol: Given: u = 10 m/s
a = - 2 m/s2
(i) Displacement, s = ut + `(1/2)` at2
Here, t = 5s.
Therefore, s = 10(5) + `(1/2)` (-2)(5)2
= 50 – 25
Therefore, after 5s, the ball will be at 25m from the starting point.
At, t = 5s, v = u + at
= 10 – 2(5)
= 0 m/s.
(ii) Displacement, when t = 10s,
S = ut + `(1/2)` at2
= 10(10) + `(1/2 )` (-2)(10)2
= 100 – 100.
Therefore after 10s, the ball will come back to the starting point.
1. A body starts from rest with uniform acceleration 2m/s2. Find the distance covered by the body in 2s.
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