Question

A car is being driven along a road at 25m s-1 when the driver suddenly notices that there is a fallen
tree blocking the road 65m ahead. The driver immediately applies the brakes giving the car a
constant deceleration of 5m s-2
. How far in front of the tree does the car come to rest?

Answer 1

Answer:

Time taken by car to stop

= 5 seconds

Step by step explanations :

Given that,

A car is being driven along a road at 25m/s

here

Initial velocity of the car = 25 m/s

given at the distance 65 meter from the car there is a fallen tree so,

the driver applies brakes

so,

final velocity will be 0 m/s.

Now we have,

Initial velocity(u) = 25 m/s

final velocity(v) = 0 m/s

distance to be travel(s) = 65 m

acceleration(a) = -5m/s²

By the equation of motion,

v = u + at

putting the values,

0 = 25 + (-5)t

-5t = -25

t = -25/-5

t = 5 s

so,

time taken by car to stop

= 5 seconds

Answer 2

Answer:

Time taken by car to come in rest

= 5 seconds

Step by step explanations :

Given that,

Initial velocity of car = 25 m/s

final velocity(v) = 0 m/s

distance to be travel(s) = 65 m

acceleration(a) = -5m/s²

By the equation of motion,

v = u + at

0 = 25 - 5t

-5t = -25

t = -25/-5

t = 5 s

so,

time taken by car to come in rest

= 5 second