Question

A car starts from rest, attains a velocity of $18km{h}^{-1}$ with an acceleration of $0.5m{s}^{-2}$, travels $4km$ with this uniform velocity and then comes to halt with a uniform deceleration of $0.2m{s}^{-2}$. The total time of travel of the car is

• A. $853s$
• B. $800s$
• C. $855s$
• D. $835s$

Answer 1

Answer: (D)

Let the car be accelerated from $A$ to $B$, it moves with uniform velocity from $B$ to $C$ of $4km$ distance and then moves with uniform deceleration of $0.2m{s}^{-2}$ from $C$ to $D$ as shown below.

For motion of car from $A$ to $B,a=0.5m{s}^{-2}$
$u=0$ and $v=18km{h}^{-1}$
$=18\times \frac{5}{18}m{s}^{-1}=5m{s}^{-1}$
Time, $t_1=\frac{v-u}{a}$.....(i)
Substituting given values of $v,u$ and $a$ for $A$ to $B$ motion, we get
$t_1=\frac{5-0}{0.5}=10s$.....(ii)
For motion of car from $B$ to $C$,
$s=4km=4000m$ and $v=5m{s}^{-1}$
$t_2=\frac{\text{ distance }}{\text{ velocity }}=\frac{4000}{5}=800s$......(iii)
For motion of car from $C$ to $D,v=0,u=5m{s}^{-1}$
and $a=-0.2m{s}^{-2}$ (negative sign shows deceleration)
Time taken, $t_3=\frac{v-u}{a}=\frac{0-5}{-0.2}=\frac{-5}{-0.2}=25s$.....(iv)
Total time taken, $T=t_1+t_2+t_3$
Substituting values of $t_1,t_2$ and $t_3$ from Eqs. (ii), (iii) and (iv) respectively, we get
$T=(10+800+25)s=835s$
Thus, total time of travel of the car is $835s$.