A bus travelling at 15ms\(^{-1}\) accelerates uniformly at 4ms\(^{-2}\), what is the distance covered in 10s?

Vkook

7 years ago

v=15m/s, a=4m/s and t=10s
Using 2nd eqn of motion,
S=ut + 1/2at^2
S=15*10+0.5*4*10^2
S=350
Therefore, the answer is C

Greatemmanuel01

5 years ago

S = ut + ½at²
S = 15 x 10 + ½ x 4 x 10²
S = 150 + 200
S = 350m

AyisaEmmanuel1

2 years ago

the parameters given were
u=15ms-¹
a=4ms-¹
s=?
t=10s

formula to use id the newtons second lar of motion......


s=ut+½at²
s=15×10+0.5×4×10²
s=150+0.5×4×100
s=150+200
s=350m

Sterix10

10 years ago

please sumone shud explain how dis question was solved

Adima2oo8

4 years ago

Please someone should further explain how the equation was solved