Math Help - Kinematics

1. Kinematics

Stuck on some problems again

- A car starts from rest and moves in a straight line. After t seconds, its acceleration, a m s^-2 is given by a=(0.2)(18-t). Find the distance travelled in the first 30 s given that the speed of the car at t=30 is 18m/s.

I tried to integrate a twice to get S but I can't get the answer of 720m.
$s = \int v\ dt = 1.8t^2 - \dfrac{1}{30}t^3 + C$
Am i doing anything wrong or missing out anything?

- A particle, travelling in a straight line, passes a fixed point O on the line with a speed of 0.5 m/s. The acceleration, a m/s^2 , of the partiicle , t seconds after passing O, is given by a = 1.4-0.6t

Find the total distance travelled by the particle between t = 0 and t = 10

For this I have tried to do the same as the above
$s = \int v\ dt = 0.7t^2 - 0.15t^3 + 0.5t + C$
and i figured that S = 0 when t = 0(since t starts increasing after particle passes O) so C = 0.

Then i found S at t=5(where it comes to inst. rest) to be 1.25m and S at t=10 to be -75m but doing the math it comes nowhere near the answer of 40m.

I appreciate any inputs

2. Originally Posted by arccos
Stuck on some problems again

- A car starts from rest and moves in a straight line. After t seconds, its acceleration, a m s^-2 is given by a=(0.2)(18-t). Find the distance travelled in the first 30 s given that the speed of the car at t=30 is 18m/s.

I tried to integrate a twice to get S but I can't get the answer of 720m.
$s = \int v\ dt = 1.8t^2 - \dfrac{1}{30}t^3 + C$
Am i doing anything wrong or missing out anything?

- A particle, travelling in a straight line, passes a fixed point O on the line with a speed of 0.5 m/s. The acceleration, a m/s^2 , of the partiicle , t seconds after passing O, is given by a = 1.4-0.6t

Find the total distance travelled by the particle between t = 0 and t = 10

For this I have tried to do the same as the above
$s = \int v\ dt = 0.7t^2 - 0.15t^3 + 0.5t + C$
and i figured that S = 0 when t = 0(since t starts increasing after particle passes O) so C = 0.

Then i found S at t=5(where it comes to inst. rest) to be 1.25m and S at t=10 to be -75m but doing the math it comes nowhere near the answer of 40m.

I appreciate any inputs
For your first problem, you are given the acceleration a as a function of time ie a=-0.2t+3.6. Integrate this twice wrt t to get the distance and sub in 30s. This will give you 720m.

3. Originally Posted by bugatti79
For your first problem, you are given the acceleration a as a function of time ie a=-0.2t+3.6. Integrate this twice wrt t to get the distance and sub in 30s. This will give you 720m.
Calculation error for the first question. My bad! Any idea how to do the second qn?

4. Originally Posted by arccos
Stuck on some problems again

- A car starts from rest and moves in a straight line. After t seconds, its acceleration, a m s^-2 is given by a=(0.2)(18-t). Find the distance travelled in the first 30 s given that the speed of the car at t=30 is 18m/s.

I tried to integrate a twice to get S but I can't get the answer of 720m.
$s = \int v\ dt = 1.8t^2 - \dfrac{1}{30}t^3 + C$
Am i doing anything wrong or missing out anything?

- A particle, travelling in a straight line, passes a fixed point O on the line with a speed of 0.5 m/s. The acceleration, a m/s^2 , of the partiicle , t seconds after passing O, is given by a = 1.4-0.6t

Find the total distance travelled by the particle between t = 0 and t = 10

For this I have tried to do the same as the above
$s = \int v\ dt = 0.7t^2 - 0.15t^3 + 0.5t + C$
what i got was $0.7t^2-0.1t^3+0.5t+C$. please recheck..
and i figured that S = 0 when t = 0(since t starts increasing after particle passes O) so C = 0.

Then i found S at t=5(where it comes to inst. rest) to be 1.25m and S at t=10 to be -75m but doing the math it comes nowhere near the answer of 40m.

I appreciate any inputs
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