Question

A Partical travels 10m in first 5 sec ,10m in 3 sec.Assuming constant acceleration what is the distance travelled in next 2 sec

Answer 1

for first 5 sec S5= 10m

s= ut +1/2 at2

10= 5u +1/2 x a x 5 x 5

2u+5a= 4 ------- eq 1

for distance traveled in first 8 sec S8= 20m

s= ut +1/2 at2

20 = 8u +1/2 x a x 8 x 8

2u + 8a == 5 ------- eq 2

solve both equations

2u = 4-- 5a

4-5a + 8a = 5

4-3a=5

a=1/3 m/s2

similarly u= 7/6

distance travelled by particle in 10 sec

S10 = u x 10 + 1/2 x a x 10 x 10

substitute value of a and u

S10=28.4

the distance in the last 2 sec == S10--S8

28.4-- 20 == 8.3m