Question

A car leaving a stop sign accelerates...what is the average speed of the car, in feet per second, equation d=1.1t^2 average speed
between t=2 and t=5?

Answer 1

Answer:

10.27 feet/sec.

Step-by-step explanation:

The distance traveled by the car, d in feet, is given by the function $d = 1.1t^{2}$ .......... (1), where t is the time in seconds.

So, at t= 2 sec, from equation (1) we get d = 1.1 × 4 = 4.4 feet.

Again, at t = 5 sec, we get d = 1.1 × 32 = 35.2 feet.

Now, average speed of the car from t = 2 sec. to t = 5 sec. will be give by

$\frac{\textrm {Total distance traveled}}{\textrm {Total time spent}}$

= $\frac{35.2 - 4.4}{5 - 2} = 10.27$ feet/sec. (Answer)