A particle starts with an initial velocity 2.5 m/s along the positive x-direction and it accelerates uniformly at the rate 0.5 m/s2.
(i) Find the distance travelled by it in the first two seconds.
(ii) how much time does it take to attain the velocity 7.5 m/s ?
(iii) How much distance will it cover in attaining the velocity 7.5 m/s?
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acc=0.5
time=2 sec
s=ut+1/2at2
=2.5×2+1/2×0.5×2×2
=5+1
=6metres
final velocity=7.5
a=v-u/t
0.5=7.5-2.5/ t
0.5t=5
t=5/0.5
t=10 sec
s=ut+1/2at2
=2.5×10+1/2×0.5×10×10
=25+25
=50 metres
Answer:
(I). The distance traveled by it in the first two seconds is 6 m.
(II). It takes to attain the velocity 7.5 m/s in 10 sec.
(III). The distance covered in 10 sec is 50 m.
Explanation:
Given that,
Initial velocity = 2.5 m/s
Acceleration = 0.5 m/s²
(I). The distance traveled by it in the first two seconds,
Using second equation of motion,
(II). We need to calculate the time,
Using equation of motion
Put the value in equation
(III). After 10 sec it attains a speed of 7.5 m/s.
So, the distance covered in 10 sec
Using equation of motion
Hence, (I). The distance traveled by it in the first two seconds is 6 m.
(II). It takes to attain the velocity 7.5 m/s in 10 sec.
(III). The distance covered in 10 sec is 50 m.