Question

A train traveling at 7.55m/s accelerates to a distance of 975m in 31.5 s. Whats the final velocity of the train?

A)227m/s
B)54.4
C)7.61
D)69.5

Answer 1

B. 54.4 m/s

Step-by-step explanation:

Using the equation;

S = ut + 1/2at²

Where;

v is the final velocity

u is the initial velocity,

a is the acceleration,

s is the displacement

Therefore;

975 =(7.55×31.5) + (0.5×a× 31.5²)

975 = 237.825 + 496.125a

496.125a = 737.175

a = 1.486 m/s²

But;

v = u + at

= 7.55 m/s + (1.486 × 31.5)

= 54.355 m/s

= 54.4 m/s

Therefore;

The final velocity is 54.4 m/s