Question

A truck traveling at a constant speed of 40.0 km/h applies its brakes and comes to a complete stop in 5.0 s.

Answer 1

Answer:

a) $v = 11.111\,\frac{m}{s}$, b) See attachment below, c) $a = - 2.222\,\frac{m}{s^{2}}$. It means that speed is decreasing at an average rate of 2.222 m/s², d) See attachment below, e) $\Delta t = 13.889\,s$, f) $\Delta s = 77.159\,m$.

Explanation:

a) The constant speed of the truck is:

$v = \left(40\,\frac{km}{h} \right)\cdot \left(\frac{1\,h}{3600\,s} \right)\cdot \left(\frac{1000\,m}{1\,km} \right)$

$v = 11.111\,\frac{m}{s}$

b) A simple graph is presented as attachment.

c) The average acceleration for the truck is:

$a = \frac{0\,\frac{m}{s} - 11.111\,\frac{m}{s} }{5\,s - 0\,s}$

$a = - 2.222\,\frac{m}{s^{2}}$

It means that speed is decreasing at an average rate of 2.222 m/s².

d) The new graph is present as attachment.

e) The time needed to regain the original speed is:

$\Delta t = \frac{11.111\,\frac{m}{s} }{0.8\,\frac{m}{s^{2}} }$

$\Delta t = 13.889\,s$

f) The distance travelled by the truck is:

$\Delta s = \frac{(11.111\,\frac{m}{s} )^{2}}{2\cdot (0.80\,\frac{m}{s^{2}} )}$

$\Delta s = 77.159\,m$

Answer 2

Answer:

Part a)

$v = 11.11 m/s$

Part c)

$a = -2.22 m/s^2$

This mean the truck is decelerating and its speed is decreasing

Part e)

$t = 13.89 s$

Part f)

$d = 77.14 m$

Explanation:

Part a)

Speed of the truck is given as

$v = 40 km/h$

as we know that

1 km = 1000 m

1 h = 3600 s

so we will have

$v =40 \times \frac{1000}{3600} m/s$

$v = 11.11 m/s$

Part c)

Average acceleration is given as

$a = \frac{v_f - v_i}{t}$

now we have

$a = \frac{0 - 11.11}{5}$

$a = -2.22 m/s^2$

Part e)

as we know that it is uniform acceleration

so we can say

$v_f - v_i = at$

$11.11 - 0 = 0.80 t$

$t = 13.89 s$

Part f)

distance traveled by the truck is given as

$d = \frac{1}{2}at^2$

$d = \frac{1}{2}(0.80)(13.89^2)$

$d = 77.14 m$