Question

A scooterist sees a bus 1 km ahead of him moving with a velocity of 10 m/s. With what speed the scooterist should move so as to overtake the bus in 100 s.
A. 10 m/s
B. 20 m/s
C. 50 m/s
D 30 m/s

Answer 1

Hint Since we know speed is to distance divided by time taken. We will use the relation to find the answer. Distance travelled by scooterists to overtake a bus is 1 km plus distance moved by bus in 100 s.

Step by step solution
According the question
Firstly, we will find the distance travelled by bus in 100 sec i.e. ${d_1}$
Let s be the speed of bus, given by
$
  \because s = \dfrac{{{d_1}}}{t} \\
  \therefore {d_1} = s*t \\
  {d_1} = 10*100 = 1000m \\
 $
Now we will find distance travelled by scooterist to overtake bus i.e.
$
  d = {d_1} + 1000 \\
  d = 1000 + 1000 \\
  d = 2000m \\
 $
Let S be the required speed to overtake the bus in 100 sec
Hence to overtake bus in 100 sec speed of scooterist should be
$S = \dfrac{d}{t} = \dfrac{{2000}}{{100}} = 20m{s^{ - 1}}$

Hence option B is correct.

Note We can also solve this problem using the concept of relative velocity.
Given, t = 100s,
s = 1000m;
v = 10m/s.
But this is the relative speed with respect to the bus

${V_s}$ = actual velocity of scooter;
${V_b}$ = velocity of bus;
$v$ = relative velocity of scooter with respect to bus.

$v = {V_s} - {V_b}$ as both bus and scooter are moving in same direction

\[\therefore 10 = {V_s} - 10\]
Hence, ${V_b} = 20m{s^{ - 1}}$

But this is the relative speed with respect to the bus

${V_s}$ = actual velocity of scooter;
${V_b}$ = velocity of bus;
$v$ = relative velocity of scooter with respect to bus.

$v = {V_s} - {V_b}$ as both bus and scooter are moving in same direction

\[\therefore 10 = {V_s} - 10\]
Hence, ${V_b} = 20m{s^{ - 1}}$