Question

a car travelling at constant velocity passes a stationary motorcycle at a traffic light as the car overtakes the motorcycle motorcycle accelerates uniformly from rest for 10 second the following displacement time graph represents the motion of both vehicles from the traffic lights and


(1) use the graph to find the magnitude of the velocity of the car (2)


how long in second it will take the motorcycle to catch f with the car (3)


how far behind the car will be motorcycle after 10 second​

Answer 1

Explanation:

1) magnitude of velocity of car

v = slope of stright x,-t graph

= ∆x/∆t

=

= (100-0)/(10-0)

= 100/10

= 10 m/s

2) Time taken by motorbike to catch car :

t = time where both curves meet

= 15 s

3) distance between car and motorcycle :

s = distance covered by car - dist covered by

motorcycle in 10 s

= 100 - 75

= 25 m

After 10 s motor cycle is 25m behind the car

Answer 2

Given:

• Graph is provided

To find:

• Velocity of car
• Time in which cycle catches car
• Difference in position at 10 seconds

Solution:

Velocity of car is :

$v = \dfrac{x2 - x1}{t2 - t1}$

$\implies v = \dfrac{150 - 100}{15 - 10}$

$\implies v = \dfrac{50}{5}$

$\implies v = 10 \: m {s}^{ - 1}$

So, velocity of car is 10 m/s.

• Now, the addition of the motorcycle and the car is same at x = 150 metres and time t = 15 seconds.

So, motorcycle catches up with the car at 15 seconds.

• At t = 10 seconds, the graph shows that the car is at x = 100 m and the cycle is at x = 75 m.

• So, the difference in position is 100 - 75 = 25 metres.

So , the cycle is 25 metres behind the car.