I think it might be A. I’m sorry if I’m wrong
Answer:
i)-6.25m/s
ii)18 metres
iii)26.5 m/s or 95.4 km/hr
Explanation:
Firstly convert 90km/hr to m/s
90 × 1000/3600 = 25m/s
(i) Apply v^2 = u^2 + 2As...where v(0m/s) is the final speed and u(25m/s) is initial speed and also s is the distance moved through(50 metres)
0 = (25)^2 + 2A(50)
0 = 625 + 100A....then moved the other value to one
-625 = 100A
Hence A = -6.25m/s^2(where the negative just tells us that its deceleration)
(ii) Firstly convert 54km/hr to m/s
In which this is 54 × 1000/3600 = 15m/s
then apply the same formula as that in (i)
0 = (15)^2 + 2(-6.25)s
-225 = -12.5s
Hence the stopping distance = 18metres
(iii) Apply the same formula and always remember that the deceleration values is the same throughout this question
0 = u^2 + 2(-6.25)(56)
u^2 = 700
Hence the speed that the car was travelling at is the,square root of 700 = 26.5m/s
In km/hr....26.5 × 3600/1000 = 95.4 km/hr
The distance from Buzău to Iaşi is 72 km. Find out the time a vehicle travels this distance if it has a speed of 72 km per hour. Expresses the speed in the unit of measurement in SI. I give the crown! Thank you
Answer:
3600seconds
Explanation:
Given parameters:
Distance = 72km
Speed = 72km/hr
Unknown:
Time it takes for the vehicle to travel the distance = ?
Solution:
Speed is the distance divided by time;
Speed =
time =
Insert the parameters;
time = = 1hr
The SI unit of time is in seconds;
3600 seconds is 1hr
The time is 3600seconds
Answer:
v_max = (1/6)e^-1 a
Explanation:
You have the following equation for the instantaneous speed of a particle:
(1)
To find the expression for the maximum speed in terms of the acceleration "a", you first derivative v(t) respect to time t:
(2)
where you have use the derivative of a product.
Next, you equal the expression (2) to zero in order to calculate t:
For t = 1/6 you obtain the maximum speed.
Then, you replace that value of t in the expression (1):
hence, the maximum speed is v_max = ((1/6)e^-1)a
It is the very last answer