(a) A car is moving with a velocity of 10ms\(^{-1}\) It then accelerates at 0.2ms\(^{-2}\) for 100m. Find, correct to two decimal places the time taken by the car to cover the distance.
(b) A particle moves along a straight line such that its distance S metres from a fixed point O is given by S = t\(^2\) - 5t + 6, where t is the time in seconds. Find its:
(a) We can use the formula:
distance = initial velocity × time + 0.5 × acceleration × time\(^2\)
To solve for time, we can rearrange this equation to:
time = [(-initial velocity) ± sqrt((initial velocity)\(^2\) + 2 × acceleration × distance)] / acceleration
Plugging in the given values, we have:
distance = 100m
initial velocity = 10ms\(^{-1}\)
acceleration = 0.2ms\(^{-2}\)
So, time = [(-10) ± sqrt((10)\(^2\) + 2 × 0.2 × 100)] / 0.2
Simplifying, we get:
time = 155.8s or -45.8s
Since time can't be negative, we discard the negative solution and conclude that the time taken by the car to cover the distance is approximately 155.8 seconds.
(b) To find the initial velocity, we need to differentiate the equation for S with respect to time (t):
S = t\(^2\) - 5t + 6
dS/dt = 2t - 5
The initial velocity is the value of dS/dt when t = 0. So, plugging in t = 0, we get:
initial velocity = dS/dt (t = 0) = 2(0) - 5 = -5 m/s
To find the distance when the particle is momentarily at rest, we need to find the time when the velocity is zero.
Setting dS/dt = 0, we get:
2t - 5 = 0
t = 2.5 s
So, the distance when the particle is momentarily at rest is:
S (t = 2.5) = (2.5)\(^2\) - 5(2.5) + 6 = 1.25 m
Therefore, the initial velocity is -5 m/s and the distance when the particle is momentarily at rest is 1.25 m.