Question

a truck Travels at a uniform velocity of 60 MS -1 for 10 second and is brought to rest in 5 second calculate the retardation of the car and the distance travelled in 15 second​

Answer 1

6m/s^2 is the retardation and 900 m

is the answer of the question

Answer 2

Answer:

a = -12m/s² and s = 15m.

Given:

initial velocity = 60 m/s

final velocity = 0 m/s

time period = 5 second

Step-by-step explanation:

First, Calculate the retardation,

Retardation = change in velocity / time taken

a = ( v - u ) / t

so,

a = ( 0 - 60 ) / 5

a = -60/5

a = -12 $ms^{-2}$

Now calculate distance traveled,

according to laws of motion.

$s = ut + \frac{1}{2}t^{2}$

u = 60 and t = 5

⇒ s = 60 × 5 + 1/2×(-12) (5)²

⇒ s = 300 - 6 × 25

⇒ s = 300 - 150

⇒ s = 150 m.

Hence, retardation is -12m/s² and distance traveled before stopping is 150 m.

#SPJ2