Question

A body moving in a straight line with an initial velocity u and a constant acceleration a covers a distance of 40 m in the fourth second and a distance of 60 metre in VI second the values of U and a r

Answer 1

Using equation, v=U+at

Here V1=40m. ,t1=4sec

V2=60m,. t2=6 sec

Also U and a value remains unchanged

so,. 40=U+4a_____(1)

60=U+6a______(2)

equating and solving equation(1) and (2)

we get, 2a=20

so a=10m/sec²

put value of a in equation(1)

we get U=0m/sec.

Answer 2

Answer:

U =10 m/s and a = 0 m/s^2

Explanation:

For t= 4 sec

40 = U (4) + 1/2 a( 4)^2

40 = 4 U + 8 a

10 = U + 2a    ---------- 1

for 6 sec

60 = U(6) + 1/2 a(6)^2

60 = 6U + 18 a

10 = U + 3a     ---------2

equation 2-1

10 =U+ 3a - (10 =U+ 2)

a= 0

putting the value of a in equation 1 we get

10= U + 2 (0)

U= 10

hence, U= 10 m/s and a = 0 m/s^2