Question

a particle is moving with a uniform acceleration.its velocity after 5s is 25 metre per second and afrer 8sis 34metre per second.the distance travelled in 12 th second is?

Answer 1

V = u + at
25= u+ 5a
34 =u + 8a
Solving these equations
9= 3a
a= 3
u= 10
S = 120+ 216- (110+ 363/2)
S= 336- 110 - 181.5
S= 44.5m

Answer 2

Answer:

$2 \times 8 + 6 - 5 + 3 \times 2 \sqrt[44 \times 3y - 4x = {2 \times 4}^{2} \times \frac{?}{?} ]{?} 4y - 5x + 57 - 1 { \\ }^{2} \cos( \beta \infty \pi \csc( log_{ \gamma 025y1y + \times {2. > 0x)89358 \times \frac{?}{?} }^{2} }(?) ) )$

solve this we will get answer