Question

A loaded truck is going at a speed Of 36km/hr and a car starts at a point 3 km behind the truck with an acceleration of 2 m/s^2 . When will the car over take the truck ?What is the distance covered by the car before it overtakes the truck

Answer 1

Answer:

60 sec

3.6 Km

Explanation:

Given that the truck is going at a speed of 36 km/hr = $36\times \frac{5}{18}$ = 10 m/sec

Given that the car starts at a point 3km=3000 m  behind the truck

Initial velocity (u) of the car = 0 m/sec

Acceleration (a) of the car = 2 m/$sec^2$

We know when they meet let the distance travelled by the truck be d Then the distance travelled by the car is d + 3000

For truck

$speed=\frac{distance}{time}$

$10 =\frac{d}{t}$

d = 10t

For car

$distance = ut +\frac{1}{2}at^2$

d+3000 = $\frac{1}{2}2t^2$

10t + 3000 = $t^2$

On solving we get t = 60 sec

The car will overtake the truck at 60 sec

The distance travelled by car is $\frac{1}{2}2t^2$ = 3600 m= 3.6 Km