A vehicle of mass 1200 kg accelerates uniformly from rest to a velocity of
15 m/s in 5 seconds. It continues at this velocity for 15 seconds before decelerating uniformly at a rate of 5m/s2 to rest.
Determine the following:
(i)The acceleration
(ii)The time taken to decelerate from 15 m/s to rest
(iii)The total distance travelled
1. V = Vo + a*t.
15 = 0 + a*5 a = 3 m/s^2.
2. V = Vo + a*t.
0 = 15 - 5*t, t = 3 s.
3. d1 = Vo*t + 0.5a*t^2.
Vo = 0, t = 5, a = 3 m/s^2, d1 = ?.
d2 = Vo*t. Vo = 15 m/s, t = 15 s,
d2 = ?.
V^2 = Vo^2 + 2a*d3.
V = 0, Vo = 15 m/s, a = -5 m/s^2, d3 = ?.
d1+d2+d3 = Total dist.
To determine the answers to these questions, we can use the equations of motion.
(i) The acceleration:
We can use the equation of motion: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. From the given information, the initial velocity (u) is 0 m/s, the final velocity (v) is 15 m/s, and the time (t) is 5 seconds. Rearranging the equation, we have a = (v - u) / t. Plugging in the values, we get a = (15 m/s - 0 m/s) / 5 s = 3 m/s^2. Therefore, the acceleration is 3 m/s^2.
(ii) The time taken to decelerate from 15 m/s to rest:
To find the time taken to decelerate, we need to use the equation of motion v = u + at, again. Here, the initial velocity (u) is 15 m/s, the final velocity (v) is 0 m/s, and the acceleration (a) is -5 m/s^2 (negative because it is a deceleration). Rearranging the equation, we have t = (v - u) / a. Plugging in the values, we get t = (0 m/s - 15 m/s) / -5 m/s^2 = 3 seconds. Therefore, the time taken to decelerate from 15 m/s to rest is 3 seconds.
(iii) The total distance traveled:
To find the total distance traveled, we need to calculate the distance covered in each phase of acceleration, constant velocity, and deceleration, and then sum them up.
Distance covered during acceleration:
Using the equation of motion s = ut + (1/2)at^2, where s is the distance, u is the initial velocity, a is the acceleration, and t is the time, we can find the distance covered during the acceleration phase. Here, the initial velocity (u) is 0 m/s, the acceleration (a) is 3 m/s^2, and the time (t) is 5 seconds. Plugging in the values, we get s = (0 m/s)*(5 s) + (1/2)*(3 m/s^2)*(5 s)^2 = 37.5 m.
Distance covered during constant velocity:
The vehicle continues at a constant velocity of 15 m/s for 15 seconds. The distance covered during this time is given by s = v * t = (15 m/s) * (15 s) = 225 m.
Distance covered during deceleration:
Using the same equation of motion s = ut + (1/2)at^2, we can find the distance covered during the deceleration phase. Here, the initial velocity (u) is 15 m/s, the final velocity (v) is 0 m/s, and the acceleration (a) is -5 m/s^2. Rearranging the equation, we have s = ut + (1/2)at^2. Plugging in the values, we get s = (15 m/s) * (3 s) + (1/2) * (-5 m/s^2) * (3 s)^2 = 67.5 m.
Therefore, the total distance traveled is 37.5 m + 225 m + 67.5 m = 330 m.