Question

If a bus accelerates uniformly from rest to attain a speed of 144 km/h in 20 sec, it covers a distance of

• A. 200 m
• B. 400 m
• C. 600 m
• D. 800 m

Answer 1

The correct option is B

400 m

Step 1: Given data
The bus is initially at rest so initial velocity U=0 m/s

Final velocity V= 144 km/h = 40 m/s

Time taken t= 20 sec

Step 2: Calculation of distance

From the 1^st Kinematic equation,

$v=u+at$

Where a is the acceleration.

Substituting the given values,

$40=0+a\left(20\right)$

40 = 20 a

a = $\frac{40}{20}$ = $2\frac{m}{s^2}$

Now, Consider 3^rd kinematic equation.

S = ut + $\frac{1}{2}$at²

where S is the distance.

S = 0( 20) + $\frac{1}{2}$ $\times 2\times {20}^2$

S= 20²

S= 400 m

Hence, Distance covered by the bus is 400 m. Hence option (b) is correct.