PRACTICE PROBLEMS SPEED VELOCITY AND ACCELERATION

Question 1 :

A missile fired from ground level rises x meters vertically upwards in t seconds and

x  =   100t - 25t²/2

Find

(i) the initial velocity of the missile

(ii) the time when the height of the missile is a maximum

(iii) the maximum height reached

(iv) the velocity with which the missile strikes the ground.

Solution :

Here x is the distance and t is time.

The distance covered by the missile in "t" seconds

x  =   100t - 25t²/2

(i) The initial velocity of the missile

Initial velocity is velocity at t  =  0.

dx/dt  =  100(1) - 25(2t)/2

=  100-25t

When t  =  0, to find the initial velocity

=  100 - 25 (0)

=  100 m/sec

So, the initial velocity is 100 m/sec.

(ii) the time when the height of the missile is a maximum

When the missile reaches its maximum height, its velocity will be 0. 

that is dx/dt  =  0

100-25t  =  0

t  =  100/25

t  =  4 seconds

The missile reaches its maximum height at 4 seconds.

(iii)  the maximum height reached

Since the missile reaches its maximum height at t  =  4

x   =  400 - 25(8)

=  400-200

=  200 meter

So, the maximum height reached is 200 m.

(iv) the velocity with which the missile strikes the ground

When the missile strikes the ground, the distance will be zero.

100t-25 t²/2  =  0

(200t-25t²)/2  =  0

200t-25t²  =  0

25t(8-t)  =  0

t  =  0, t  =  8

t = 0 is not possible. So, let us take t = 8

Velocity at t  =  8

dx/dt  =  100-25t

=  100-25(8)

=  100-200

=  -100 meter/second

Problem 2 :

A particle of unit mass moves so that displacement after t seconds is given by

x  =  3cos (2t-4)

Find the acceleration and kinetic energy at the end of 2 seconds. (K.E = (1/2) m v²)

Solution :

Here x represents the displacement of the object and t displacement of a particle in seconds.

x(t)  =  3cos(2t-4)

mass   = 1

to find acceleration, we have to different the given equation two times

velocity dx/dt  =  3[-sin (2t-4)] (2(1) - 0)

v   =  - 6 sin (2t-4)

velocity at t = 2

v  =  -6 sin (2(2) - 4)

=  -6 sin (4-4)

=  -6 sin (0)

v  =  0

Acceleration d²x/dt²  =  -6 cos (2t-4) (2(1)- 0)

=  -6cos (2t-4) 2

=  -12cos (2t-4)

Plug t  =  2

=  -12cos (2(2)-4)

=  -12cos (0)

= -12

Kinetic energy K.E  =  (1/2) m v²               

=  (1/2) (1) (0)²

=  (1/2) (0)

=  0

Problem 3 :

The distance x meters traveled by a vehicle in time t seconds after the brakes are applied is given by

x  =  20t-(5/3)t²

Determine

(i) the speed of the vehicle (in km/hr) at the instant the brakes are applied and

(ii) the distance the car traveled before it stops.

Solution :

the distance x meters traveled by a vehicle in time t seconds

x  =  20t - (5/3)t²

To find the speed of the vehicle, we find the value of dx/dt for that we have to differentiate the given function with respect to t.

dx/dt  =  20(1) - (5/3)(2t)

=  20-(10 t/3)

the speed of the vehicle (in km/hr) at the instant the brakes are applied 

t  =  0

=  20-(10(0)/3)

=  20 - 0

velocity at t = 0   

=  20 m/sec

to find velocity after the brakes are applied

=  20 meter/seconds

to convert this into km/hr we have to multiply it by the fraction 3600/1000

=  (20 x 3600)/1000

=  72 km/hr

Problem 4 :

Newton's law of cooling is given by θ  =  θ₀° e^(⁻kt), where the excess of temperature at zero time is θ₀° C and at time t seconds is θ° C. Determine the rate of change of temperature after 40 s, given that θ₀ = 16° C and k = -0.03.(e^1.2  =  3.3201)

Solution :

Newton's law of cooling θ = θ₀° e^(⁻kt)

θ₀°  =  16° C

k  =  -0.03

rate of change of temperature with respect to time

dθ/dt  =  -k θ₀° e^(⁻kt)

t  =  40

dθ/dt  =  -(-0.03) (16)e^(⁻0.03) (40)

=   0.48  e^(⁻1.2)

=  0.48(3.3201)

=   1.5936° C/s

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