Question

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

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

Answer 1

Answer: (D)

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

For motion of car from $A$ to $B, a=0.5\, ms ^{-2}$
$u=0 $ and $v =18 \,km \,h ^{-1} $
$=18 \times \frac{5}{18} ms ^{-1}=5 \,ms ^{-1}$
Time, $ t_{1}=\frac{v-u}{a}$.....(i)
Substituting given values of $v, u$ and $a$ for $A$ to $B$ motion, we get
$t_{1}=\frac{5-0}{0.5}=10 \,s$.....(ii)
For motion of car from $B$ to $C$,
$s =4 \,km =4000 \,m $ and $ v=5 \,ms ^{-1} $
$t_{2} =\frac{\text { distance }}{\text { velocity }}=\frac{4000}{5}=800 \,s$......(iii)
For motion of car from $C$ to $D, v=0, u=5 \,ms ^{-1}$
and $a=-0.2 \,ms ^{-2}$ (negative sign shows deceleration)
Time taken, $t_{3}=\frac{v-u}{a}=\frac{0-5}{-0.2}=\frac{-5}{-0.2}=25 \, s$.....(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
$T=(10+800+25) s =835 \,s$
Thus, total time of travel of the car is $835 \, s$.