Question

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.

Answer 1

Answer:

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