Question

A particle moves with uniform acceleration covers 25m in 5th second and 35m in 7sec of its journey then distance travelled in its 3rd second will be

Answer 1

Distance covered in nth second of the motion is given by equation

$s_{n^{th}} = u + \frac{a}{2}(2n - 1)$

here

u = initial speed

a = acceleration

now we have

$25 = u + \frac{a}{2}(2*5 - 1)$

$25 = u + 4.5a$

now again we have

$35 = u + \frac{a}{2}(2*7 - 1)$

$35 = u + 6.5a$

now by above two equations

a = 5 m/s^2

u = 2.5 m/s

now distance traveled in 3rd second

$d = 2.5 + \frac{5}{2}(2*3 - 1)$

$d = 15 m$

so the total distance moved will be 15 m

Answer 2

Answer:

Distance covered in nth second of the motion is given by equation

s_{n^{th}} = u + \frac{a}{2}(2n - 1)snth=u+2a(2n−1)

here

u = initial speed

a = acceleration

now we have

25 = u + \frac{a}{2}(2*5 - 1)25=u+2a(2∗5−1)

25 = u + 4.5a25=u+4.5a

now again we have

35 = u + \frac{a}{2}(2*7 - 1)35=u+2a(2∗7−1)

35 = u + 6.5a35=u+6.5a

now by above two equations

a = 5 m/s^2

u = 2.5 m/s

now distance traveled in 3rd second

d = 2.5 + \frac{5}{2}(2*3 - 1)d=2.5+25(2∗3−1)

d = 15 md=15m

so the total distance moved will be 15 m