In a simple model of the motion of a car, its velocity vm/s, at time t seconds is given by
v= (3t^2 -4t +1)i + (5t- 3)j where t is greater than or equal to 0
a) Find the intial speed of the car.
b) Find when the car is moving parallel to the vector (i - j)
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- Thread Starter
- 15-01-2010 11:51
- 15-01-2010 11:55
omg i have no idea
u dont have enough to do suvat do you?
- 15-01-2010 11:59
Wouldnt you put t=0 to find initial speed?
And then put your v equation equal to (i-j) to find t when moving parallel?
Im just guessing =/
Oh god, ive got this soon aswell =/
- 15-01-2010 12:05
I guess it starts at t=0
i - 3j = initial velocity
Initial speed = square root 1^2 + 3^2 = square root of 10
For parallel to (i - j)
Velocity will be xi - xj
3t^2 -4t +1 = -5t +3
3t^2 + t -2 = 0
3t^2 +3t -2t -2 = 0
(3t -2) (t + 1) = 0
so T = 2/3 or t = -1