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3 years ago

9

The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hou

r (mph). At full power, the car can accelerate from zero to 29.0 mph in time 1.10 s .A more realistic car would cause the wheels to spin in a manner that would result in the ground pushing it forward with a constant force (in contrast to the constant power in Part A). If such a sports car went from zero to 29.0 mph in time 1.10 s , how long would it take to go from zero to 58.0 mph ?

1 answer:

bekas [8.4K]3 years ago

6 0

Answer:

2.2 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

Converting mph to m/s

$29\ mph=29\times 0.44704=12.96\ m/s$

$58\ mph=58\times 0.44704=25.93\ m/s$

$v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{12.96-0}{1.1}\\\Rightarrow a=11.78\ m/s^2$

Considering this acceleration to be constant

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{25.93-0}{11.78}\\\Rightarrow t=2.20\ s$

Time it would take to go from zero to 58.0 mph is 2.2 seconds

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Let's divide the motion of the sprinter in two parts:

- In the first part, he starts with velocity u = 0 and accelerates with constant acceleration $a_1$ for a total time $t_1$ During this part of the motion, he covers a distance equal to $s_1 = 45 m$ , until he finally reaches a velocity of $v_1 = u + a_1t_1$ . We can use the following suvat equation:

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$s_1 = \frac{1}{2}a_1 t_1^2$ (1)

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- In the second part, he continues with constant speed $v_1 = a_1 t_1$ , covering a distance of $d_2 = 55 m$ in a time $t_2$ . This part of the motion is a uniform motion, so we can use the equation

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We also know that the total time is 10.0 s, so

$t_1 + t_2 = 10.0 s\\t_2 = (10.0-t_1)$

Therefore substituting into the 2nd equation

$s_2 = a_1 t_1 (10-t_1)$

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$s_2 = \frac{2s_1}{t_1^2}t_1 (10-t_1)=\frac{2s_1}{t_1}(10-t_1)=\frac{20 s_1}{t_1}-2s_1$

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So from (3) we find the acceleration in the first phase:

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b) 14.5 m/s

The speed of the sprinter remains constant during the last 55 m of motion, so we can just use the suvat equation

$v_1 = u +a_1 t_1$

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$a_1 =2.34 m/s^2$ is the acceleration

$t_1 = 6.2 s$ is the time of the first part

Solving the equation,

$v_1 = 0 +(2.34)(6.2)=14.5 m/s$

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