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1. June 2001 Q7

At time t=0 a small package is projected from a point B which is 2.4m above a point A on horizontal ground. The package is projected with speed 23.75m/s at an angle @ to the horizontal, where [email protected]=4/3. The package strikes the ground at the point C, as show in Fig. 2. The package is modelled as a particle moving freely under gravity.

(a) Find the time taken for the package to reach C.

A lorry moves along the line AC, approacing A with constant speed 18m/s. At time t=0, the rear of the lorry passes A and the lorry starts to slow down. It comes to rest T seconds later. The acceleration, a m/s², of the lorry at time t seconds is given by a=-0.25t², 0 <= t <= T.

(b) Find the speed of the lorry at time t seconds.

(c) Hence show that T = 6.

(d) Show that when the package reaches C it is just under 10m behind the rear of the moving lorry.
-----

I did a, b and c, but I just can't seem to do d.
The answers to a and b are:
(a) 4s
(b) 18 - (1/12)t³

My method for d is:
Package: S = Ux * t = [email protected] * 4 = 57m
Lorry:
t = 6, u = 18, a = -0.25at²
S = ut + 0.5at² = 18t -(0.125)t^4 = -54m (!)
2. (Original post by shift3)
June 2001 Q7

At time t=0 a small package is projected from a point B which is 2.4m above a point A on horizontal ground. The package is projected with speed 23.75m/s at an angle @ to the horizontal, where [email protected]=4/3. The package strikes the ground at the point C, as show in Fig. 2. The package is modelled as a particle moving freely under gravity.

(a) Find the time taken for the package to reach C.

A lorry moves along the line AC, approacing A with constant speed 18m/s. At time t=0, the rear of the lorry passes A and the lorry starts to slow down. It comes to rest T seconds later. The acceleration, a m/s², of the lorry at time t seconds is given by a=-0.25t², 0 <= t <= T.

(b) Find the speed of the lorry at time t seconds.

(c) Hence show that T = 6.

(d) Show that when the package reaches C it is just under 10m behind the rear of the moving lorry.
-----

I did a, b and c, but I just can't seem to do d.
The answers to a and b are:
(a) 4s
(b) 18 - (1/12)t³

My method for d is:
Package: S = Ux * t = [email protected] * 4 = 57m
Lorry:
t = 6, u = 18, a = -0.25at²
S = ut + 0.5at² = 18t -(0.125)t^4 = -54m (!)
are you sure you've posted that correctly? SUVAT equations only work when acceleration is constant, and in your question, the acceleration increases with time. so your SUVAT equation won't work.
3. s(package) = 57m
============

For lorry,
v = 18 -(1/12)t³
ds/dt = v = 18 -(1/12)t³
s = ∫ v dt = 18t - (1/48)^4 + C

at t=0, s = 0 => c = 0

s = 18t - (1/48)^4

at t=4,
s = 18.4 - (1/48)4^4
s = 72 - 5.33
s = 66.67 m
========

which is just less than 10m from where the package hits (57m)

Edit: fixed a silly arithemetic mistake.
4. (Original post by 4Ed)
are you sure you've posted that correctly? SUVAT equations only work when acceleration is constant, and in your question, the acceleration increases with time. so your SUVAT equation won't work.
Hmmm... How come I didn't spot that?

Here's the solution, if anyone's interested:
v = 18 - (1/12)t³
[integrate wrt t]
s = 18t - (1/48)t^4 + c
When s=0, t=0, so c=0.
i.e. s(t) = 18t - (1/48)t^4
So, s(4) =~ 66.7 (3sf)

66.7 - 57 = 9.7 < 10, as required.

Thanks.

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Updated: June 24, 2004
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