A car travels with uniform acceleration and travels 12m in 2nd second of its motion and cover
20m 4 second of its motion. Calculate its acceleration and also the distance covered in 3
second of its motion.
:) 15.5 m
a) 14 m
c) 18 m
b) 16 in
Answers
The acceleration of the car is 4 m/s² and also the distance covered in 3 seconds of its motion is 16 m.
Explanation:
We know that for uniform acceleration, the formula to find the distance covered by the car in nth second is given by,
Sn = u + [(½) * a(2n-1)]
For 2nd second of motion:
Here the distance covered by the car = 12 m
∴ S₂ = u + [(½) * a(2*2-1)]
⇒ 12 = u + 3a/2
⇒ 24 = 2u + 3a …… (i)
For 4th second of motion:
Here the distance covered by the car = 20 m
∴ S₄ = u + [(½) * a(2*4-1)]
⇒ 20 = u + 7a/2
⇒ 40 = 2u + 7a …… (ii)
Now, by subtracting (i) from (ii), we get
16 = 4a
⇒ a = 4 m/s²
Substituting a= 4 m/s² in (i), we get
24 = 2u + 3*4
⇒ 2u = 24 – 12
⇒ u = 12/2 = 6 m/s
Thus,
The distance “S3” covered in 3rd second of its motion is given by,
= u + ½ * a[2*3-1]
= 6 + ½ * 4[6-1]
= 16 m
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