Question

2. A particle is moving under a constant acceleration of 3 m/s". The initial
velocity of the particle is 10 m/s. Calculate the distance travelled by the
particle in 5 s and also in the 5th second of its motion.​

Answer 1

Answer:

12.5

Explanation:

s=ut+1/2at^2

s=o-0.5*25

s=12.5

Answer 2

Given, a= 3m/s^2 (Acceleration)
(Initial velocity) U= 10m/s
(Time) T= 5s
To find - s (distance)
Solution,
Using 2nd equation of motion,
V= ut+ 1/2at^2
V= (10)(5) + 1/2 (3) (5)^2
V= 50+ 75/2
V= 100/2+75/2
V= 175/2
V= 87.5 m/s.
Now, using the third equation of motion,
V^2= u^2 + 2as
(87.5)^2 = (10)^2 + 2(3)(s)
7656.25= 100 + 6s
7656.25-100= 6s
7556.25=6s
s= 7556.25/6
s= 1259.38m (approx.)
Hence, the distance travelled by the particle is 1259.38 m.