Question

can anyone tell me where I can find the answer or can anyone answer this please?


I'll mark you as BRAINLIEST, friend please let me know the answer​

Answer 1

Answer:

Initial speed, v=72 km/h

⟹v=  

3600

72×10  

3

 

​  

 

⟹v=20 m/s

Distance, 2r=50 cm

⟹r=25×10  

−2

 m

Angular displacement, θ=20×2π

⟹θ=40π rad

Initial angular velocity, ω=  

r

v

​  

 

⟹ω=  

25×10  

−2

 

20

​  

 

⟹ω=  

5

4

​  

×10  

2

 

⟹ω=80 rad/s

Final angular velocity, ω  

′

=0 rad/s

From kinematics:

ω  

′2

−ω  

2

=2αθ (where α is angular acceleration)

⟹0−6400=2α×40π

⟹α=  

80π

−6400

​  

 

⟹α=−25.5 rad/s  

2

Answer 2

Answer:

According to the problem, speed of the car as well as truck = 72 km/hf car as well as truck = 72 km/h

= $72*\frac{5}{18} m/s = 20m/s$

The time required to stop the truck= 5$s$

Finally, the truck comes to rest, so the final velocity of truck will be zero.

Retardation produced by truck:

$v=u+a_{1}t$

$0=20+a_{1}*5$

or $a_{1}= -4 m/s^{2}$

The time required to stop the car 3$s$

Finally, the car comes to rest just behind the truck in the same time to avoid the collision, so the final velocity of car will also be zero.

Retardation produced by car is

$v= u+a_{c}t$

$0=20+a_{c}*3$

or $a_{c}=-\frac{20}{3}m/s^{2}$

Let the car be at a distance s from the truck when the truck gives signal and $t$ be the time taken to cover the distance.

As human response time is 0.5, in this time the car will cover some distance with uniform velocity. There, time of retarded motion of the car is $(t-0.5)$ $s.$

The velocity of the car after the time $t$,

$v_{c}= u-at = 20 - (\frac{20}{3} )(t-0.5)$

The velocity of the truck after the time $t$,

$v_{1} =20-4t$

To avoid the car bump onto the track,

$20-\frac{20}{3}(t-0.5)=20-4t$

$4t=\frac{20}{3}(t-0.5)$

⇒$t=\frac{2.5}{2} = \frac{5}{4}s$

Distance traveled by truck in time $t$,

$s_{t}=u_{t}t+\frac{1}{2}a_{t}t^{2}$

⇒ $s_{t}= 20*\frac{5}{4}+\frac{1}{2} *(-4)*(\frac{5}{4})^{2}=21.875m$

Distance traveled by car in time $t$ = Distance traveled by car in $0.5s$ ( without retardation) + distance traveled by car in $(t-0.5)$ $s$ (with retardation)

$s_{c}=(20*0.5)+20(\frac{5}{4}-0.5)-\frac{1}{2}(\frac{20}{3})(\frac{5}{4}-0.5)^{2} = 23.125m$

∴ $s_{c}-s_{t}=23.125-21.875=1.250m$

Therefore, to avoid the collision with the truck, the car must maintain a distance from the truck more than 1.250 m.

Hope it helps, please mark me as brainliest, thanks! :)