# Question #774be

Jul 27, 2017

a) It does not come to a halt
b) ${V}_{f i n a l}$=1.88 $m \cdot {s}^{- 1}$

#### Explanation:

a)
${p}_{i}$=15600 $N \cdot s$

$i m p u l s e$= $\left(860 \cdot 16\right) N \cdot {s}^{1}$

$i m p u l s e$= $13760 N \cdot {s}^{1}$

Therefore, the car won't stop.

b)

${p}_{n e t}$ = 15600-13760
${p}_{n e t}$ = $1840 N \cdot {s}^{1}$
$m$ = 980 kg

${v}_{f} = \frac{{p}_{n e t}}{m}$

${v}_{f} = \frac{1840}{980}$

${v}_{f} = 1.88 m \cdot {s}^{- 1}$

Jul 27, 2017

(a) No
(b) $1.88 m {s}^{-} 1$, rounded to two decimal places.

#### Explanation:

Momentum of car $| \vec{p} | = 15600 N s$
Mass of car $= 980 k g$
Velocity of car $= | \vec{p} \frac{|}{m} = \frac{15600}{980} = \frac{780}{49} m {s}^{-} 1$

Stopping force applied $= 860 N$
(it is assumed that it is applied against the direction of initial momentum)
Retardation$= | \vec{F} \frac{|}{m} = \frac{860}{980} = \frac{43}{49} m {s}^{-} 2$

using the kinematic expression

$v = u + a t$

final velocity after $16 s$
$v = \frac{780}{49} + \left(- \frac{43}{49} \times 16\right)$
$v = \frac{780}{49} - \frac{688}{49}$
$v = 1.88 m {s}^{-} 1$, rounded to two decimal places.