A rocket sled for testing equipment under
large accelerations starts at rest and acceler-
ates according to the expression
a= (1.5 m/s^3)t+ (2.9 m/s^2).
How far does the sled move in the time
interval t= 0 to t= 2.4 s?

Question

A rocket sled for testing equipment under
large accelerations starts at rest and acceler-
ates according to the expression
a= (1.5 m/s^3)t+ (2.9 m/s^2).
How far does the sled move in the time
interval t= 0 to t= 2.4 s?

anonymous · Answer

the integral of acceleration is velocity, then the integral of velocity is position, sound familiar?

anonymous · Answer

$$velocity = \int\limits_{ }^{ } acceleration = \int\limits_{ }^{ } 1.5t + 2.9 = 0.75t^{2} + 2.9t$$

anonymous · Answer

can you do the integral of 0.75t^2 + 2.9t to get the position equation, which then we can just plug in the time and get the answer?