Question

A particle moving along a straight line with a constant acceleration of - 4 m per second square passes through a point on the line with a velocity of + 8 metre per second at the moment when the distance travelled by the particle in 6 second in 5 seconds

Answer 1

Given, acceleration , a = -4 m/s²
Intial velocity , u = 8 m/s
Here, acceleration is negative sign which indicates particle is retarding motion.
Let after t time particle will be rest.
∴ final velocity , v = 0 m/s
Now, use formula , v = u + at
0 = 8 - 4t
t = 2 sec
hence, velocity of particle is decreases at t = 2 sec after that particle moves in direction of Acceleration .
so, distance covered in 6 sec = distance travelled in first 2 sec + distance travelled by last 4sec
= | ut₁ + 1/2 at₁² | + | 0.t₂ + 1/2 at₂² |
In last 4 sec , initial velocity will take zero. Because particle will be rest after two sec.
Here, t₁ = 2 and t₂ = 4
Now, Total distance travelled = 8 × 2 + 1/2 × (-4) × 2² + |1/2(-4) × (4)²|
= 16 - 8 + |-32|
= 8 + 32 = 40 m

Similarly , total distance travelled in 5 sec = distance travelled in 2 sec + distance travelled in last 3 sec
= 16 - 8 + | -18|
= 8 + 18 = 26 m

Answer 2

uy=10sin 60=53√m/s

⇒     t=2uyg=2×53√10=3√s

Sx=uxt+12axt2

1.15=5×t−12a×t2

1.15=5×3√−32a

 3a2=5×1.73−1.15=8.65−1.15

 3a2=7.5  

 ⇒   a=153=5m/s2

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