Kinematics

Jt061952 SG

I need help with part (iii).

A particle moving in a straight line passes a fixed point A with a velocity of -7 m/s. The acceleration after passing A is given by a = 8 - 2t for 0 <= t <= T. On reaching its max speed at T secs, the particle begins to decelerate at a constant rate of 1.5 m/s2. until it comes to rest at point B.

i) Show that the particle first comes to instantaneous rest when t=1.

able to do this one, INTEG ( a = 8 - 2t) to get v = 8t - t^2 - 7

ii) Find the max speed attained by the particle and the value T.

able to do this one,

0 = 8 - 2t, t = 4 sec

v = 8(t) - (4)(4) - 7 = 9 m/s

iii) Find the total distance travelled by the particle from A to B.

t^2 - 8t +7 =0 , so I have (t - 7)(t - 1) = 0

Paticle come to rest at B (7sec??)

So I INTEG v = t^2 - 8t +7 from (0 to 1) and (1 to 7).

Not sure about this one. Can someone help? Thanks.

Answer is 48 1/3 m.

Mikethm SG

Part iii) is wrong. Because upon reaching max speed, the particle does not follow the original behaviour (aka equations).

"A particle moving in a straight line passes a fixed point A with a velocity of -7 m/s. The acceleration after passing A is given by a = 8 - 2t for 0 <= t <= T. On reaching its max speed at T secs, the particle begins to decelerate at a constant rate of 1.5 m/s2. until it comes to rest at point B."

You reach max speed when t=4. So the original equations applies for 0<=t<=4 only

Integ v = t^2 - 8t +7 from (0 to 1) and (1 to 4). to find distance travelled in 1st 4 seconds.

9/1.5 = 6s

Particle come to rest 6s after t=4

Distance travelled from t=4 to t=10

= 0.5 x 6 x 9

= 27m

Add 27m to your distance travelled in 1st 4 seconds and you should get the 48 1/3m.

Note that I never bother to find distance travelled in 1st 4s as you are capable of doing so.

Jt061952 SG

Hi Mikethm - thanks, appreciate your answer given. jt