Answer to Question #163226 in Mechanics | Relativity for M

Question

A motorboat starting from a dead stop accelerates at a constant 1.7 m/s² for 2.5 s, then very rapidly roars up to 4.4 m/s² and holds it constant for 4.0 s. What is its average acceleration over the first 4.9 s of motion? 

 

What is its speed after 6.5 s? 

 

Next the boat decelerates to a stop in 19.0 s at a constant rate. What is the boat's acceleration during this last portion of the trip? 

Answer 1

$\text{On each segment of movement, the speed is calculated}$

$\text{by the formula for uniformly accelerated movement}$

$v (t)= v_0 +at$

$\text{for }t\in [0,2.5]$

$v (t)= v_0 +a_1t$

$v_0 = 0;a_1=1.7$

$v (t)= 1.7t$

$\text{for }t\in [2.5,6.5]$

$v (t)= v_0 +a_2(t-2.5)$

$v_0=v(2.5)= 1.7*2.5 = 4.25$

$a_2=4.4$

$v (t)= 4.25 +4.4(t-2.5)\ (1)$

$a(4.9) = \frac{v(4.9)-v(0)}{\Delta{t}}=\frac{v(4.9)-v(0)}{4.9}$

$v(4.9)= 4.25+4.4*(4.9-2.5) = 14.81 \text{ of } (1)$

$v(0)=0$

$a(4.9)=\frac{14.81}{4.9}\approx3.02$

$v(6.5)= 4.25+4.4(6.5-2.5)=21.85 \text{ of }(1)$

$\text{for }t\in [6.5,25.5]$

$v (t)= v_0 +a_3(t-6.5)$

$v_0=v(6.5)= 21.85$

$v (t)= 21.85 +a_3(t-6.5)$

$v(19+6.5) = 0\text{( next the boat decelerates to a stop in 19.0 s )}$

$v(19+6.5) = 21.85+a_3(19+6.5-6.5)$

$21.85+19a_3=0$

$a_3= -1.15$

$\text{Answer: }a(4.9)\approx3.02m/s^2;v(6.5)=21.85m/s;a_3=-1.15m/s^2$