Science, asked by aaron22, 1 year ago

(a) The velocity-time graph of a car is given below. The car weighs 1000 kg. (I) What is the distance travelled by the car in the first 2 seconds? (ii) What is the breaking force at the end of 5 seconds to bring the car to a stop within one second. (b) Derive the equation s=ut+1/2at square using graphical method.

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Answers

Answered by ssms1013
23
(a)
velocity = distance / time 
(1) distance = velocity * time = 15 * 2 = 30 m 

(b)
distance = area of figure ABCFE
=area of triangle ABE + area of square BCFE
= 1/2 * at*t + ut
=>
s = ut + 1/2 at^2
Answered by sonuvuce
1

(i) The distance travelled by the car is first 2 seconds is 15 m

(ii) The braking force at the end of 5 seconds to bring the car to a stop within one second is 15000 N

Explanation:

Given mass of car m = 1000 kg

(a) We know that in a velocity time graph, the distance travelled in time t = area under the velocity-time graph upto time t

Therefore, the distance travelled by the car in first 2 seconds

= Area of ΔABE

=\frac{1}{2}\times BE\times AE

=\frac{1}{2}\times 15\times 2

=15 m

(b) From graph it is clear that the car moves with a constant acceleration from A to B and then with a zero acceleration from B to C and then with negative acceleration from C to D

Acceleration at the end of 5 seconds

= Acceleration (a) from  C to D = Slope of line CD

                                                = \frac{15-0}{5-6}

                                                = =-15 m/s²

Therefore the breaking force at the end of 5 seconds

F=ma

\implies F=1000\times (-15)

\implies F=-15000 N

(negative sign indicates that force applied is in the opposite direction of motion)

(b) Distance travelled from C to D

s = Area of ΔCFD

=\frac{1}{2}\times CF\times FD</p><p>[tex]=\frac{1}{2}\times 15\times (6-5)

=\frac{15}{2} m

We have calculated, acceleration a from velocity C to velocity D as -15 m/s²

Initial velocity at C, u = 15 m/s

Time interval t = 6-5 = 1 seconds

Therefore,

ut+\frac{1}{2}at^2

=15\times 1+\frac{1}{2}(-15)\times 1^2

=15-\frac{15}{2}

=\frac{15}{2}

=s

Hope this answer is helpful.

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