Answer to Question #149045 in Mechanics | Relativity for Melanie Belandres

Question

5. A soapbox derby race car starts at rest and moves 300 m down a track in 22.4 s. Calculate its acceleration (assumed constant) and its speed at the end of the track.

6. A car traveling at 35 mi/hr is stopped in a distance of 2.5 ft when it hits a tree. (a) Calculate the average deceleration of the car in m/s2. (b) Calculate the time needed to stop the car (assuming constant deceleration).

7. An automobile and a truck start from rest at the same instant, with the automobile initially at some distance behind the truck. The truck has a constant acceleration of 2 m/s2 and the automobile an acceleration of 3 m/s2. The automobile overtakes the truck after the truck has moved 75 m.
a) How long does it take the automobile to overtake the truck?
b) How far was the car behind the truck initially?
c) What is the velocity of each when they are abreast?

Answer 1

Explanations & Calculations

5)

• Since the acceleration is constant any of the 4 motion equations can be used & by using appropriately,
• $\qquad\qquad \begin{aligned} \small s&=ut+\frac{1}{2}at^2\\ \small 300&= \small 0+\frac{1}{2}a(22.4)^2\\ a&= \small \bold{1.196\,ms^{-2}} \end{aligned}$ : Since s & t are known
• $\qquad\qquad \begin{aligned} \small s&=\frac{(v+u)t}{2}\\ \small 300&= \small \frac{(v+0)\times 22.4}{2}\\ \small v&= \small \bold{26.786\,ms^{-1}} \end{aligned}$ : Can be calculated using other equations as a is known by then

6)

• Again as a constant deceleration, any of the 4 motion equations can be used.
• $\small 35mi/hr=\large\frac{35\times 1.609\times 1000m}{(3600)s}=\small 56315ms^{-1}$
• $\small 2.5ft=2.5\times 0.3048m=0.762m$

• (a) using $\small v^2 = u^2 +2as$

$\qquad\qquad \begin{aligned} \small 0^2&= \small 56315^2+2a\times 0.762\\ \small a&= \small -2.08\times 10^{9}ms^{-2} \end{aligned}$

• (b)using $\small s = \large\frac{(v+u)t}{2}$

$\qquad\qquad \begin{aligned} \small 0.762 &= \small \frac{(0+56315)t}{2}\\ \small t &= \small \bold{2.706\times 10^{-5}s} \end{aligned}$

7)

• Both the vehicles spend the same time until they meet. Therefore, truck travels 75m during that period while the a.m travels $\small (x+75) m$. (if the initial separation between them is x)
• (a)Then apply $\small s=ut +\frac{1}{2}at^2$ for the motion of the truck.

$\qquad\qquad \begin{aligned} \small 75&= \small 0+\frac{1}{2}\times 2ms^{-2}\times t^2\\ \small t&= \small \pm\sqrt{\frac{75\times 2}{2}}\\ \small t&= \small \bold{8.66s} \end{aligned}$

• (b) By applying $\small s = ut +\frac{1}{2}at^2$ for the automobile, for its motion until the overtake

$\qquad\qquad \begin{aligned} \small x+75 &= \small 0+\frac{1}{2}\times 3ms^{-2}\times (8.66s)^2\\ \small x &= \small \bold{37.49m} \end{aligned}$

• (c) By applying $\small v = u+at$ on both the vehicles

$\qquad\qquad \begin{aligned} \end{aligned}$For the a.m For the truck

$\qquad\qquad \begin{aligned} \small v_a&= \small 0+3ms^{-2}\times 8.66s\\ &=\small \bold{25.98ms^{-1}} \end{aligned}$ $\begin{aligned} \small v_t&= \small 0+2ms^{-2}\times 8.66s\\ &= \small \bold{17.32ms^{-1}} \end{aligned}$