A rocket sled for testing equipment under large accelerations starts at rest and accelerates according to the expression a = (3.6 m/s 3) t + (5.6 m/s 2). How far does the sled move in the time interval t = 0 to t = 1.6 s? Answer in units of m.

9.6 m

Explanation:

This is a case of motion under variable acceleration. So no law of motion formula will be applicable here. We shall have to integrate the given equation.

a = 3.6 t + 5.6

d²x / dt² = 3.6 t + 5.6

Integrating on both sides

dx / dt = 3.6 t² / 2 + 5.6 t + c

where c is a constant.

dx / dt = 1.8 t² + 5.6 t + c

when t = 0, velocity dx / dt is zero

Putting these values in the equation above

0 = 0 + 0 + c

c = 0

dx / dt = 1.8 t² + 5.6 t

Again integrating on both sides

x = 1.8 t³ / 3 + 5.6 x t² / 2 + c₁

x = 0.6 t³ + 2.8 t² + c₁

when t = 0, x = 0

c₁ = 0

x = 0.6 t³ + 2.8 t²

when t = 1.6

x =.6 x 1.6³ + 2.8 x 1.6²

= 2.4576 + 7.168

= 9.6256

9.6 m